#include <bits/stdc++.h>
using namespace std;
long long n, t;
long long cnt[60][60];
int main() {
  scanf("%lld%lld", &n, &t);
  for (int i = 0; i < 60; i++) {
    cnt[i][0] = 1;
    for (int j = 1; j <= i; j++) {
      cnt[i][j] = cnt[i - 1][j] + cnt[i - 1][j - 1];
    }
  }
  int want = 0;
  while (t > 1) {
    want++;
    if (t % 2) {
      printf("0\n");
      return 0;
    }
    t /= 2;
  }
  long long ans = 0;
  for (int i = 1; n > 0 && want >= 0; i++) {
    if ((1ull << i) > n) {
      while ((1ull << i) > n) i--;
      n -= (1ull << i);
      ans += cnt[i][want];
      want--;
    } else {
      n -= (1ull << i);
      ans += cnt[i][want];
    }
  }
  printf("%lld\n", ans);
  return 0;
}

### Prompt

Develop a solution in cpp to the problem described below:

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 n, t;
long long cnt[60][60];
int main() {
  scanf("%lld%lld", &n, &t);
  for (int i = 0; i < 60; i++) {
    cnt[i][0] = 1;
    for (int j = 1; j <= i; j++) {
      cnt[i][j] = cnt[i - 1][j] + cnt[i - 1][j - 1];
    }
  }
  int want = 0;
  while (t > 1) {
    want++;
    if (t % 2) {
      printf("0\n");
      return 0;
    }
    t /= 2;
  }
  long long ans = 0;
  for (int i = 1; n > 0 && want >= 0; i++) {
    if ((1ull << i) > n) {
      while ((1ull << i) > n) i--;
      n -= (1ull << i);
      ans += cnt[i][want];
      want--;
    } else {
      n -= (1ull << i);
      ans += cnt[i][want];
    }
  }
  printf("%lld\n", ans);
  return 0;
}

```