#include <bits/stdc++.h>
using namespace std;
long long cnk[50][50] = {1};
long long calc(long long mask, long long mx, int cnt, int n) {
  if (mask > mx) return 0;
  if (cnt == 0) return 1;
  if (cnt > n + 1) return 0;
  if (mask + (1LL << (n + 1)) - 1 <= mx) return cnk[n + 1][cnt];
  return calc(mask + (1LL << n), mx, cnt - 1, n - 1) +
         calc(mask, mx, cnt, n - 1);
}
bool solve() {
  long long n, t;
  if (!(cin >> n >> t)) return false;
  if ((t - 1) & t) {
    printf("0\n");
    return true;
  }
  int cnt = 0;
  for (; t; t /= 2, cnt++)
    ;
  for (int i = 1; i < 50; ++i)
    for (int j = 0; j <= i; ++j) {
      cnk[i][j] = 0;
      if (j > 0) cnk[i][j] += cnk[i - 1][j - 1];
      if (j < i) cnk[i][j] += cnk[i - 1][j];
    }
  long long ans = calc(0, n + 1, cnt, 60);
  if (cnt == 1) ans--;
  cout << ans << endl;
  return true;
}
int main() {
  solve();
  return 0;
}

### Prompt

Please formulate 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 cnk[50][50] = {1};
long long calc(long long mask, long long mx, int cnt, int n) {
  if (mask > mx) return 0;
  if (cnt == 0) return 1;
  if (cnt > n + 1) return 0;
  if (mask + (1LL << (n + 1)) - 1 <= mx) return cnk[n + 1][cnt];
  return calc(mask + (1LL << n), mx, cnt - 1, n - 1) +
         calc(mask, mx, cnt, n - 1);
}
bool solve() {
  long long n, t;
  if (!(cin >> n >> t)) return false;
  if ((t - 1) & t) {
    printf("0\n");
    return true;
  }
  int cnt = 0;
  for (; t; t /= 2, cnt++)
    ;
  for (int i = 1; i < 50; ++i)
    for (int j = 0; j <= i; ++j) {
      cnk[i][j] = 0;
      if (j > 0) cnk[i][j] += cnk[i - 1][j - 1];
      if (j < i) cnk[i][j] += cnk[i - 1][j];
    }
  long long ans = calc(0, n + 1, cnt, 60);
  if (cnt == 1) ans--;
  cout << ans << endl;
  return true;
}
int main() {
  solve();
  return 0;
}

```