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Get random double (floating point) value from random byte array between 0 and 1 in C#?

Assume I have an array of bytes which are truly random (e.g. captured from an entropy source).
byte[] myTrulyRandomBytes = MyEntropyHardwareEngine.GetBytes(8);
Now, I want to get a random double precision floating point value, but between the values of 0 and positive 1 (like the Random.NextDouble() function performs).
Simply passing an array of 8 random bytes into BitConverter.ToDouble() can yield strange results, but most importantly, the results will almost never be less than 1.
I am fine with bit-manipulation, but the formatting of floating point numbers has always been mysterious to me. I tried many combinations of bits to apply randomness to and always ended up finding the numbers were either just over 1, always VERY close to 0, or very large.
Can someone explain which bits should be made random in a double in order to make it random within the range 0 and 1?

Though working answers have been given, I'll give an other one, that looks worse but isn't:
long asLong = BitConverter.ToInt64(myTrulyRandomBytes, 0);
double number = (double)(asLong & long.MaxValue) / long.MaxValue;
The issue with casting from an ulong to double is that it's not directly supported by hardware, so it compiles to this:
vxorps xmm0,xmm0,xmm0
vcvtsi2sd xmm0,xmm0,rcx ; interpret ulong as long and convert it to double
test rcx,rcx ; add fixup if it was "negative"
jge 000000000000001D
vaddsd xmm0,xmm0,mmword ptr [00000060h]
vdivsd xmm0,xmm0,mmword ptr [00000068h]
Whereas with my suggestion it will compile more nicely:
vxorps xmm0,xmm0,xmm0
vcvtsi2sd xmm0,xmm0,rcx
vdivsd xmm0,xmm0,mmword ptr [00000060h]
Both tested with the x64 JIT in .NET 4, but this applies in general, there just isn't a nice way to convert an ulong to a double.
Don't worry about the bit of entropy being lost: there are only 262 doubles between 0.0 and 1.0 in the first place, and most of the smaller doubles cannot be chosen so the number of possible results is even less.
Note that this as well as the presented ulong examples can result in exactly 1.0 and distribute the values with slightly differing gaps between adjacent results because they don't divide by a power of two. You can change them exclude 1.0 and get a slightly more uniform spacing (but see the first plot below, there is a bunch of different gaps, but this way it is very regular) like this:
long asLong = BitConverter.ToInt64(myTrulyRandomBytes, 0);
double number = (double)(asLong & long.MaxValue) / ((double)long.MaxValue + 1);
As a really nice bonus, you can now change the division to a multiplication (powers of two usually have inverses)
long asLong = BitConverter.ToInt64(myTrulyRandomBytes, 0);
double number = (double)(asLong & long.MaxValue) * 1.08420217248550443400745280086994171142578125E-19;
Same idea for ulong, if you really want to use that.
Since you also seemed interested specifically in how to do it with double-bits trickery, I can show that too.
Because of the whole significand/exponent deal, it can't really be done in a super direct way (just reinterpreting the bits and that's it), mainly because choosing the exponent uniformly spells trouble (with a uniform exponent, the numbers are necessarily clumped preferentially near 0 since most exponents are there).
But if the exponent is fixed, it's easy to make a double that's uniform in that region. That cannot be 0 to 1 because that spans a lot of exponents, but it can be 1 to 2 and then we can subtract 1.
So first mask away the bits that won't be part of the significand:
x &= (1L << 52) - 1;
Put in the exponent (1.0 - 2.0 range, excluding 2)
x |= 0x3ff0000000000000;
Reinterpret and adjust for the offset of 1:
return BitConverter.Int64BitsToDouble(x) - 1;
Should be pretty fast, too. An unfortunate side effect is that this time it really does cost a bit of entropy, because there are only 52 but there could have been 53. This way always leaves the least significant bit zero (the implicit bit steals a bit).
There were some concerns about the distributions, which I will address now.
The approach of choosing a random (u)long and dividing it by the maximum value clearly has a uniformly chosen (u)long, and what happens after that is actually interesting. The result can justifiably be called a uniform distribution, but if you look at it as a discrete distribution (which it actually is) it looks (qualitatively) like this: (all examples for minifloats)
Ignore the "thicker" lines and wider gaps, that's just the histogram being funny. These plots used division by a power of two, so there is no spacing problem in reality, it's only plotted strangely.
Top is what happens when you use too many bits, as happens when dividing a complete (u)long by its max value. This gives the lower floats a better resolution, but lots of different (u)longs get mapped onto the same float in the higher regions. That's not necessarily a bad thing, if you "zoom out" the density is the same everywhere.
The bottom is what happens when the resolution is limited to the worst case (0.5 to 1.0 region) everywhere, which you can do by limiting the number of bits first and then doing the "scale the integer" deal. My second suggesting with the bit hacks does not achieve this, it's limited to half that resolution.
For what it's worth, NextDouble in System.Random scales a non-negative int into the 0.0 .. 1.0 range. The resolution of that is obviously a lot lower than it could be. It also uses an int that cannot be int.MaxValue and therefore scales by approximately 1/(231-1) (cannot be represented by a double, so slightly rounded), so there are actually 33 slightly different gaps between adjacent possible results, though the majority of the gaps is the same distance.
Since int.MaxValue is small compared to what can be brute-forced these days, you can easily generate all possible results of NextDouble and examine them, for example I ran this:
const double scale = 4.6566128752458E-10;
double prev = 0;
Dictionary<long, int> hist = new Dictionary<long, int>();
for (int i = 0; i < int.MaxValue; i++)
{
long bits = BitConverter.DoubleToInt64Bits(i * scale - prev);
if (!hist.ContainsKey(bits))
hist[bits] = 1;
else
hist[bits]++;
prev = i * scale;
if ((i & 0xFFFFFF) == 0)
Console.WriteLine("{0:0.00}%", 100.0 * i / int.MaxValue);
}

This is easier than you think; its all about scaling (also true when going from a 0-1 range to some other range).
Basically, if you know that you have 64 truly random bits (8 bytes) then just do this:
double zeroToOneDouble = (double)(BitConverter.ToUInt64(bytes) / (decimal)ulong.MaxValue);
The trouble with this kind of algorithm comes when your "random" bits aren't actually uniformally random. That's when you need a specialized algorithm, such as a Mersenne Twister.

I don't know wether it's the best solution for this, but it should do the job:
ulong asLong = BitConverter.ToUInt64(myTrulyRandomBytes, 0);
double number = (double)asLong / ulong.MaxValue;
All I'm doing is converting the byte array to a ulong which is then divided by it's max value, so that the result is between 0 and 1.

To make sure the long value is within the range from 0 to 1, you can apply the following mask:
long longValue = BitConverter.ToInt64(myTrulyRandomBytes, 0);
longValue &= 0x3fefffffffffffff;
The resulting value is guaranteed to lay in the range [0, 1).
Remark. The 0x3fefffffffffffff value is very-very close to 1 and will be printed as 1, but it is really a bit less than 1.
If you want to make the generated values greater, you could set a number higher bits of an exponent to 1. For instance:
longValue |= 0x03c00000000000000;
Summarizing: example on dotnetfiddle.

If you care about the quality of the random numbers generated, be very suspicious of the answers that have appeared so far.
Those answers that use Int64BitsToDouble directly will definitely have problems with NaNs and infinities. For example, 0x7ff0000000000001, a perfectly good random bit pattern, converts to NaN (and so do thousands of others).
Those that try to convert to a ulong and then scale, or convert to a double after ensuring that various bit-pattern constraints are met, won't have NaN problems, but they are very likely to have distributional problems. Representable floating point numbers are not distributed uniformly over (0, 1), so any scheme that randomly picks among all representable values will not produce values with the required uniformity.
To be safe, just use ToInt32 and use that int as a seed for Random. (To be extra safe, reject 0.) This won't be as fast as the other schemes, but it will be much safer. A lot of research and effort has gone into making RNGs good in ways that are not immediately obvious.

Simple piece of code to print the bits out for you.
for (double i = 0; i < 1.0; i+=0.05)
{
var doubleToInt64Bits = BitConverter.DoubleToInt64Bits(i);
Console.WriteLine("{0}:\t{1}", i, Convert.ToString(doubleToInt64Bits, 2));
}
0.05: 11111110101001100110011001100110011001100110011001100110011010
0.1: 11111110111001100110011001100110011001100110011001100110011010
0.15: 11111111000011001100110011001100110011001100110011001100110100
0.2: 11111111001001100110011001100110011001100110011001100110011010
0.25: 11111111010000000000000000000000000000000000000000000000000000
0.3: 11111111010011001100110011001100110011001100110011001100110011
0.35: 11111111010110011001100110011001100110011001100110011001100110
0.4: 11111111011001100110011001100110011001100110011001100110011001
0.45: 11111111011100110011001100110011001100110011001100110011001100
0.5: 11111111011111111111111111111111111111111111111111111111111111
0.55: 11111111100001100110011001100110011001100110011001100110011001
0.6: 11111111100011001100110011001100110011001100110011001100110011
0.65: 11111111100100110011001100110011001100110011001100110011001101
0.7: 11111111100110011001100110011001100110011001100110011001100111
0.75: 11111111101000000000000000000000000000000000000000000000000001
0.8: 11111111101001100110011001100110011001100110011001100110011011
0.85: 11111111101011001100110011001100110011001100110011001100110101
0.9: 11111111101100110011001100110011001100110011001100110011001111
0.95: 11111111101110011001100110011001100110011001100110011001101001

Approximation of n points to the curve with the best fit

I have a list of n points(2D): P1(x0,y0), P2(x1,y1), P3(x2,y2) …
Points satisfy the condition that each point has unique coordinates and also the coordinates of each point xi, yi> 0 and xi,yi are integers.
The task is to write an algorithm which make approximation of these points
to the curve y = | Acos (Bx) | with the best fit (close or equal to 100%)
and so that the coefficients A and B were as simple as possible.
I would like to write a program in C # but the biggest problem for me is to find a suitable algorithm. Has anyone would be able to help me with this?

Taking B as an independent parameter, you can solve the fitting for A using least-squares, and compute the fitting residual.
The residue function is complex, with numerous minima of different value, and an irregular behavior. Anyway, if the Xi are integer, the function is periodic, with a period related to the LCM of the Xi.
The plots below show the fitting residue for B varying from 0 to 2 and from 0 to 10, with the given sample points.

Based on How approximation search works I would try this in C++:
// (global) input data
#define _n 100
double px[_n]; // x input points
double py[_n]; // y input points
// approximation
int ix;
double e;
approx aa,ab;
// min max step recursions ErrorOfSolutionVariable
for (aa.init(-100,+100.0,10.00,3,&e);!aa.done;aa.step())
for (ab.init(-0.1,+ 0.1, 0.01,3,&e);!ab.done;ab.step())
{
for (e=0.0,ix=0;ix<_n;ix++) // test all measured points (e is cumulative error)
{
e+=fabs(fabs(aa.a*cos(ab.a*px[ix]))-py[ix]);
}
}
// here aa.a,ab.a holds the result A,B coefficients
It uses my approx class from the question linked above
you need to set the min,max and step ranges to match your datasets
can increase accuracy by increasing the recursions number
can improve performance if needed by
using not all points for less accurate recursion layers
increasing starting step (but if too big then it can invalidate result)
You should also add a plot of your input points and the output curve to see if you are close to solution. Without more info about the input points it is hard to be more specific. You can change the difference computation e to match any needed approach this is just sum of abs differences (can use least squares or what ever ...)

Fast lookup suffering from floating point inaccuracies

Suppose I have equally spaced doubles (64 bit floating point numbers) x0,x1,...,xn. Equally spaced means that for all i, x(i+1) - xi is constant; call it w for width.
Given a number y in the range [x0,xn] I want to find the largest i such that xi <= y.
A naive approach would visit each i in turn (O(n)). Marginally better is to use a binary search (O(log n)).
A constant time lookup would be to calculate (y-x0)/w and cast it to an integer. However, this will occasionally give the wrong result due to floating point inaccuracy. E.g. Suppose there are 100 intervals of width 0.01 starting at 0.
(int)(0.29/0.01) = 28 //want 29 here
Can I retain the constant time lookup but ensure that the results are always identical to the binary search? Performing the calculation with decimals rather than doubles for 'w' and 'x0' seems to work here, but will it always work? I could always follow the direct lookup with a comparison with the xs either side, but this seems ugly and inefficient.
To clarify - I am given the xi and the value y as doubles - I cannot change this. But any intermediate calculation performed before returning the integer index can use any datatypes I like. Additionally, I can perform one-off "preparation" work in order to make the runtime calculation faster.
Edit: Apologies - turns out that I didn't check "equally spaced" properly - these numbers are often not "equally spaced" when their difference is calculated using floating point arithmetic.

Do the following
Calculate (int)(0.29/0.01) = 28 //want 29 here
Next, calculate back i * 0.01 for i between 28-1 and 28+1 and pick up the one that is correct.

What do you mean equally spaced? If can make some assumptions about the numbers, for example - that they increase on an interval, you can actually use median selction that is O(1) in the best case and O(log2(N)) in the worst case.

Force a relatively small inaccuracy in floating point number

I need to add a very small value to a floating point value to make it insignificantly different so that it fails an equality test.
To avoid issues with precision, instead of adding a very small number, I have opted to add a relatively small number. Is this a good solution? or is there a reliable way to add an even smaller number?
matrix.m00 += matrix.m00 * 0.0000001f;
matrix.m11 += matrix.m11 * 0.0000001f;
matrix.m22 += matrix.m22 * 0.0000001f;
From reading I have found that the best solution is to use the next representable floating point number. Though in C# the process of doing this either a) requires unmanaged/unsafe code, or b) uses BitConverter which is too slow. So I figured that the above solution would work, but I would like to know if there are any gotchyas.

You can add an ulp to any double (depends on the double); that is the smallest number that you can add or subtract to it that will change its value.
Calculate the unit in the last place (ULP) for doubles
next higher/lower IEEE double precision number
Though, those posts all use BitConverter. I discovered a post that discusses how to add an ulp without unsafe code or BitConverter, though:
http://realtimemadness.blogspot.com/2012/06/nextafter-in-c-without-allocations-of.html

Sure there's a gotcha. If any of these values is 0, then you'll be adding exactly 0, i.e. not modifying the value at all.
Is there any reason why you couldn't use unsafe code to do this?

The minimum number you can add to a floating point number such that a different number is produced is a function of the original number, it's not some constant. Call this function Epsilon(x).
Epsilon(0), i.e. the minimum floating point number you can add to floating point 0 such that a distinguishable number is produced, can be found in the static value Double.Epsilon.
Even using a "large" epsilon like 1 will eventually fail, though. For example, this returns true in C#:
var big = 10000000000000000.0;
Console.WriteLine(big == (big + 1.0));
So unless you are sure that your input is in some fixed range of magnitude (e.g. all close to 0), you can't just fudge it with a single constant.

Calculate factorials in C#

How can you calculate large factorials using C#? Windows calculator in Win 7 overflows at Factorial (3500). As a programming and mathematical question I am interested in knowing how you can calculate factorial of a larger number (20000, may be) in C#. Any pointers?
[Edit] I just checked with a calc on Win 2k3, since I could recall doing a bigger factorial on Win 2k3. I was surprised by the way things worked out.
Calc on Win2k3 worked with even big numbers. I tried !50000 and I got an answer, 3.3473205095971448369154760940715e+213236
It was very fast while I did all this.
The main question here is not only to find out the appropriate data type, but also a bit mathematical. If I try to write a simple factorial code in C# [recursive or loop], the performance is really bad. It takes multiple seconds to get an answer. How is the calc in Windows 2k3 (or XP) able to perform such a huge factorial in less than 10 seconds? Is there any other way of calculating factorial programmatically in C#?

Have a look at the BigInteger structure:
http://msdn.microsoft.com/en-us/library/system.numerics.biginteger.aspx
Maybe this can help you implement this functionality.
CodeProject has an implementation for older versions of the framework at http://www.codeproject.com/KB/cs/biginteger.aspx.

If I try to write a simple factorial code in C# [recursive or loop], the performance is really bad. It takes multiple seconds to get an answer.
Let's do a quick order-of-magnitude calculation here for a naive implementation of factorial that performs n multiplications. Suppose we are on the last step. 19999! is about 218 bits. 20000 is about 25 bits; we'll assume that it is a 32 bit integer. The final multiplication therefore involves the addition of up to 25 partial results each roughly 218 bits long. The number of bit operations will therefore be on the order of 223.
That's for the last stage; there will be 20000 = 216 such operations at each stage, so that is a total of about 239 operations. Some of them will of course be cheaper, but we're going for an order of magnitude here.
A modern processor does about 232 operations per second. Therefore it will take about 27 seconds to get the result.
Of course, the big integer library writers were not naive; they take advantage of the ability of the chip to do many bit operations in parallel. They're probably doing the math in 32 bit chunks, giving speedups of a factor of 25. So our total order-of-magnitude calculation is that it should take about 22 seconds to get a result.
22 is 4. So your observation that it takes a few seconds to get a result is expected.
How is the calc in Windows 2k3 (or XP) able to perform such a huge factorial in less than 10 seconds?
I don't know. Extreme cleverness in exploiting the math operations on the chip probably. Or, using a non-naive algorithm for calculating factorial. Or, possibly they are using Stirling's Approximation and getting an inexact result.
Is there any other way of calculating factorial programmatically in C#?
Sure. If all you care about is the order of magnitude then you can use Stirling's Approximation. If you care about the exact value then you're going to have to compute it.

There exist sophisticated computational algorithms for efficiently computing the factorials of large, arbitrary precision numbers. The Schönhage–Strassen algorithm, for instance, allows you to perform asymptotically fast multiplication for arbitrarily large integers.
Case in point, Mathematica computes 22000! on my machine in less than 1 second. The Implementation Notes page at reference.wolfram.com states:
(Mathematica's) n! uses an O(log(n) M(n)) algorithm of Schönhage based on dynamic decomposition to prime powers.
Unfortunately, the implementation of such algorithms is both complicated and error prone. Rather than trying to roll your own implementation, it may be wiser for you to license a copy of Mathematica (or a similar product that meets your functional and performance needs) and either use it, or a .NET programming interface to it, to perform your computation.

Have you looked at System.Numerics.BigInteger?

Using System.Numerics BigInteger
var bi = new BigInteger(1);
var factorial = 171;
for (var i = 1; i <= factorial; i++)
{
bi *= i;
}
will be calculated to
1241018070217667823424840524103103992616605577501693185388951803611996075221691752992751978120487585576464959501670387052809889858690710767331242032218484364310473577889968548278290754541561964852153468318044293239598173696899657235903947616152278558180061176365108428800000000000000000000000000000000000000000
For 50000! it takes a couple seconds to calculate but it seems to work and the result is a 213237 digit number and that's also what Wolfram says.

You will probably have to implement your own arbitrary precision numeric type.
There are various approaches. probably not the most efficient, but perhaps the simplest is to have variable length arrays of byte (unsigned char). Each element represents a digit. ideally this would be included in a class, and you can then add a method which let's you multiply the number with another arbitrary precision number. A multiply with a standard C# integer would probably also be a good idea, but a little trickier to implement.

Since they don't give you the result down to the last digit, they may be "cheating" using some approximation.
Check out http://mathworld.wolfram.com/StirlingsApproximation.html
Using Stirling's formula you can calculate (an approximation of) the factorial of n in logn time. Of course, they might as well have a dictionary with pre-calculated values of factorial(n) for every n up to one million, making the calculator show the result extremely fast.

This answer covers limits for basic .Net types to compute and represent n!
Basic code to calculate factorial for "SomeType" that supports multiplication:
SomeType factorial = 1;
int n = 35;
for (int i = 1; i <= n; i++)
{
factorial *= i;
}
Limits for built in number types:
short - correct results up to 7!, incorrect results afterwards, code returns 0 starting 18 (similar to int)
int - correct results up to 12!, incorrect results afterwards, code returns 0 starting at 34 (Why computing factorial of realtively small numbers (34+) returns 0)
float - precise results up to 14!, correct but not precise afterwards, returns infinity starting at 35
long - correct results up to 20!, incorrect results afterwards, code returns 0 starting at 66 (similar to int)
double - precise results up to 22!, correct but not precise afterwards, returns infinity starting at 171
BigInteger - precise and upper limit is set by memory usage only.
Note: integer types overflow pretty quickly and start producing incorrect results. Realistically if you need factorials for any practical usage long is the type to go (up to 20!), if you can't expect limited numbers - BigInteger is the only type provided in .Net Framework to provide precise results (albeit slow for large numbers as there is no built-in optimized n! method)

You need a special big-number library for this. This link introduces the System.Numeric.BigInteger class, and incidentally has an example program that calculates factorials. But don't use the example! If you recurse like that, your stack will grow horribly. Just write a for-loop to do the multiplication.

I don't know how you could do this in a language without arbitrary precision arithmetic. I guess a start could be to count factors of 5 and 2, removing them from the product, and add on these zeroes at the end.
As you can see there are many.
>>> factorial(20000)
<<non-zeroes removed>>0000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000L