ACM for C (m, n) = m!/((m-n)!n!) The number of zeros at the end of the binary number

Description

The number of all combinations of n (n ≤ m) elements taken from m different elements is called the number of combinations of n elements taken from m different elements. The formula for calculating the number of combinations is as follows:

C (m, n) = m!/((m - n)! N!)

Now, if you write the combined number C (m, n) as a binary number, how many zeros are there at the end of the binary number?

Input

The first line is the number of test samples T, followed by the T test sample, each test sample takes a row, there are two numbers, followed by m and n, where n ≤ m≤ 1000.

Output

Outputs the number of each combined number converted to the end zero after the binary number.

Sample Input

2

4 2

1000 500

Sample Output

1

6

problem-solving ideas: This topic is to find the number of factors, m!/((m-n)!*n!) is equal to the product from the n+1 has multiplied to m divided by the factorial of M-n, that is, the number of factors in n+1 to M minus 1 to m-n number of factors.

Program code:

#include <iostream>

using namespace std;

Main ()

{

int T;

cin>>t;

 while (t--)

{

int m,n;

cin>>m>>n;

int r=0,v=0;

 for (int i=n+1;i<=m;i++)

{

int s=i;

 while (s%2==0)

{

S=S/2;

r++;

}

}

 for (int j=1;j<=m-n;j++)

{

int b=j;

 while (b%2==0)

{

B=B/2;

v++;

}

}

printf ("%d\n", r-v);

}

return 0;

}

ACM for C (m, n) = m!/((m-n)!n!) The number of zeros at the end of the binary number