#include <bits/stdc++.h>
using namespace std;
long long com[101][101];
void calccom() {
  com[0][0] = 1;
  for (int i = 1; i <= 100; i++) {
    for (int j = 0; j <= i; j++) {
      if (j == 0) {
        com[i][j] = 1;
      } else {
        if (j == i) {
          com[i][j] = 1;
        } else {
          com[i][j] = com[i - 1][j] + com[i - 1][j - 1];
        }
      }
    }
  }
}
int main() {
  calccom();
  long long gen, ko;
  scanf("%I64d%I64d", &gen, &ko);
  long long now = 1;
  int nn = -1;
  for (int i = 0;; i++) {
    if (ko == now) {
      nn = i;
      break;
    }
    if (ko < now) {
      break;
    }
    now *= 2;
  }
  if (nn == -1) {
    printf("0\n");
    return 0;
  }
  long long ans = 0;
  now = 0;
  long long pl = 1;
  int pt;
  for (pt = 0;; pt++) {
    if (now + pl <= gen + 1) {
      ans += com[pt][nn];
      now += pl;
    } else {
      break;
    }
    pl *= 2;
  }
  for (; pt >= 0; pt--) {
    if (nn < 0) {
      break;
    }
    if (now + pl <= gen + 1) {
      ans += com[pt][nn];
      now += pl;
      nn--;
    }
    pl /= 2;
  }
  if (ko == 1) {
    ans--;
  }
  printf("%I64d\n", ans);
}

### Prompt

Your task is to create a CPP solution to 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;
long long com[101][101];
void calccom() {
  com[0][0] = 1;
  for (int i = 1; i <= 100; i++) {
    for (int j = 0; j <= i; j++) {
      if (j == 0) {
        com[i][j] = 1;
      } else {
        if (j == i) {
          com[i][j] = 1;
        } else {
          com[i][j] = com[i - 1][j] + com[i - 1][j - 1];
        }
      }
    }
  }
}
int main() {
  calccom();
  long long gen, ko;
  scanf("%I64d%I64d", &gen, &ko);
  long long now = 1;
  int nn = -1;
  for (int i = 0;; i++) {
    if (ko == now) {
      nn = i;
      break;
    }
    if (ko < now) {
      break;
    }
    now *= 2;
  }
  if (nn == -1) {
    printf("0\n");
    return 0;
  }
  long long ans = 0;
  now = 0;
  long long pl = 1;
  int pt;
  for (pt = 0;; pt++) {
    if (now + pl <= gen + 1) {
      ans += com[pt][nn];
      now += pl;
    } else {
      break;
    }
    pl *= 2;
  }
  for (; pt >= 0; pt--) {
    if (nn < 0) {
      break;
    }
    if (now + pl <= gen + 1) {
      ans += com[pt][nn];
      now += pl;
      nn--;
    }
    pl /= 2;
  }
  if (ko == 1) {
    ans--;
  }
  printf("%I64d\n", ans);
}

```