#include <bits/stdc++.h>
using namespace std;
int main(int argc, char **argv) {
  long long c[50][50];
  long long n, t;
  long long res = 0;
  cin >> n >> t;
  if (t & t - 1) {
    cout << 0 << endl;
    return 0;
  }
  for (int i = 0; i <= 40; i++) {
    c[i][0] = 1;
    for (int j = 1; j <= i; j++) {
      c[i][j] = c[i - 1][j] + c[i - 1][j - 1];
    }
  }
  int p;
  for (p = 0; 1LL << p < t; p++)
    ;
  p++;
  n += 2;
  int pr = 0;
  for (int i = 40; i >= 0 && pr <= p; i--) {
    if (1LL << i & n) {
      res += c[i][p - pr];
      n -= 1LL << i;
      pr++;
    }
  }
  if (p == 1) res--;
  cout << res << endl;
  return 0;
}

### Prompt

In cpp, your task is to solve the following problem:

Maxim loves to fill in a matrix in a special manner. Here is a pseudocode of filling in a matrix of size (m + 1) × (m + 1):

<image>

Maxim asks you to count, how many numbers m (1 ≤ m ≤ n) are there, such that the sum of values in the cells in the row number m + 1 of the resulting matrix equals t.

Expression (x xor y) means applying the operation of bitwise excluding "OR" to numbers x and y. The given operation exists in all modern programming languages. For example, in languages C++ and Java it is represented by character "^", in Pascal — by "xor".

Input

A single line contains two integers n and t (1 ≤ n, t ≤ 1012, t ≤ n + 1).

Please, do not use the %lld specifier to read or write 64-bit integers in C++. It is preferred to use the cin, cout streams or the %I64d specifier.

Output

In a single line print a single integer — the answer to the problem. 

Examples

Input

1 1


Output

1


Input

3 2


Output

1


Input

3 3


Output

0


Input

1000000000000 1048576


Output

118606527258

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
int main(int argc, char **argv) {
  long long c[50][50];
  long long n, t;
  long long res = 0;
  cin >> n >> t;
  if (t & t - 1) {
    cout << 0 << endl;
    return 0;
  }
  for (int i = 0; i <= 40; i++) {
    c[i][0] = 1;
    for (int j = 1; j <= i; j++) {
      c[i][j] = c[i - 1][j] + c[i - 1][j - 1];
    }
  }
  int p;
  for (p = 0; 1LL << p < t; p++)
    ;
  p++;
  n += 2;
  int pr = 0;
  for (int i = 40; i >= 0 && pr <= p; i--) {
    if (1LL << i & n) {
      res += c[i][p - pr];
      n -= 1LL << i;
      pr++;
    }
  }
  if (p == 1) res--;
  cout << res << endl;
  return 0;
}

```