#include <bits/stdc++.h>
using namespace std;
long long C[50][50];
int main() {
  C[0][0] = 1;
  for (int i = 1; i < 50; ++i) {
    C[i][0] = C[i][i] = 1;
    for (int j = 1; j < i; ++j) C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
  }
  long long n, t;
  while (cin >> n >> t) {
    if ((t & (t - 1))) {
      puts("0");
      continue;
    }
    ++n;
    int lt = 0, ln = 0;
    for (long long x = t; x; x >>= 1) ++lt;
    for (long long x = n; x; x >>= 1) ++ln;
    int rem = lt;
    long long ans = 0;
    for (int i = ln - 1; i >= 0; --i) {
      if ((n >> i) & 1) {
        ans += C[i][rem--];
        if (rem < 0) break;
      }
    }
    if (lt == 1) --ans;
    if (rem == 0) ++ans;
    cout << ans << endl;
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution 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 C[50][50];
int main() {
  C[0][0] = 1;
  for (int i = 1; i < 50; ++i) {
    C[i][0] = C[i][i] = 1;
    for (int j = 1; j < i; ++j) C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
  }
  long long n, t;
  while (cin >> n >> t) {
    if ((t & (t - 1))) {
      puts("0");
      continue;
    }
    ++n;
    int lt = 0, ln = 0;
    for (long long x = t; x; x >>= 1) ++lt;
    for (long long x = n; x; x >>= 1) ++ln;
    int rem = lt;
    long long ans = 0;
    for (int i = ln - 1; i >= 0; --i) {
      if ((n >> i) & 1) {
        ans += C[i][rem--];
        if (rem < 0) break;
      }
    }
    if (lt == 1) --ans;
    if (rem == 0) ++ans;
    cout << ans << endl;
  }
  return 0;
}

```