#include <bits/stdc++.h>
using namespace std;
long long f[45][45];
int main() {
  long long i, j, n, t, tt;
  scanf("%I64d%I64d", &n, &t);
  for (tt = 0; t > 1; t >>= 1, tt++)
    if (t & 1) {
      printf("0\n");
      return 0;
    }
  for (i = 0; i <= 41; ++i) {
    f[i][0] = 1;
    for (j = 1; j <= i; ++j) f[i][j] = f[i - 1][j] + f[i - 1][j - 1];
  }
  long long ans = 0;
  for (i = 1; n - (1ll << i) >= 0; ++i) {
    ans += f[i][tt];
    n -= 1ll << i;
  }
  for (i--; i >= 0 && tt >= 0; --i)
    if (n >= (1ll << i)) {
      ans += f[i][tt--];
      n -= 1ll << i;
    }
  printf("%I64d\n", ans);
  return 0;
}

### Prompt

Generate 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 f[45][45];
int main() {
  long long i, j, n, t, tt;
  scanf("%I64d%I64d", &n, &t);
  for (tt = 0; t > 1; t >>= 1, tt++)
    if (t & 1) {
      printf("0\n");
      return 0;
    }
  for (i = 0; i <= 41; ++i) {
    f[i][0] = 1;
    for (j = 1; j <= i; ++j) f[i][j] = f[i - 1][j] + f[i - 1][j - 1];
  }
  long long ans = 0;
  for (i = 1; n - (1ll << i) >= 0; ++i) {
    ans += f[i][tt];
    n -= 1ll << i;
  }
  for (i--; i >= 0 && tt >= 0; --i)
    if (n >= (1ll << i)) {
      ans += f[i][tt--];
      n -= 1ll << i;
    }
  printf("%I64d\n", ans);
  return 0;
}

```