#include <bits/stdc++.h>
using namespace std;
long long ans[2][50];
long long ans2[2][50];
int main() {
  long long n, t;
  cin >> n >> t;
  if (t & (t - 1)) {
    printf("%d\n", 0);
    return 0;
  }
  n++;
  int cnt = __builtin_ctzll(t) + 1;
  ans[0][0] = 1;
  for (int i = 60; i >= 0; i--) {
    for (int j = 0; j < 45; j++) {
      ans2[1][j] += ans[1][j];
      ans2[1][j + 1] += ans[1][j];
      if (n & (1LL << i)) {
        ans2[0][j + 1] += ans[0][j];
        ans2[1][j] += ans[0][j];
      } else {
        ans2[0][j] += ans[0][j];
      }
    }
    memcpy(ans, ans2, sizeof(ans));
    memset(ans2, 0, sizeof(ans));
  }
  cout << (ans[0][cnt] + ans[1][cnt]) - (t == 1) << endl;
  return 0;
}

### Prompt

Your challenge is to write 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 ans[2][50];
long long ans2[2][50];
int main() {
  long long n, t;
  cin >> n >> t;
  if (t & (t - 1)) {
    printf("%d\n", 0);
    return 0;
  }
  n++;
  int cnt = __builtin_ctzll(t) + 1;
  ans[0][0] = 1;
  for (int i = 60; i >= 0; i--) {
    for (int j = 0; j < 45; j++) {
      ans2[1][j] += ans[1][j];
      ans2[1][j + 1] += ans[1][j];
      if (n & (1LL << i)) {
        ans2[0][j + 1] += ans[0][j];
        ans2[1][j] += ans[0][j];
      } else {
        ans2[0][j] += ans[0][j];
      }
    }
    memcpy(ans, ans2, sizeof(ans));
    memset(ans2, 0, sizeof(ans));
  }
  cout << (ans[0][cnt] + ans[1][cnt]) - (t == 1) << endl;
  return 0;
}

```