#include <bits/stdc++.h>
using namespace std;
long long dp[65][65][2];
int b[65];
int main() {
  ios::sync_with_stdio(false);
  cout.tie(0);
  cin.tie(0);
  long long n, t, cnt;
  cin >> n >> t;
  if (__builtin_popcountll(t) > 1) {
    cout << 0;
    return 0;
  }
  cnt = __builtin_ctzll(t) + 1;
  cerr << cnt << endl;
  n++;
  for (int i = 0; i < 64; i++) b[i] = (n >> (63 - i)) & 1;
  dp[0][0][0] = 1;
  for (int i = 0; i < 64; i++)
    for (int c = 0; c < 64; c++) {
      if (b[i]) {
        dp[i + 1][c][1] += dp[i][c][0];
        dp[i + 1][c + 1][0] += dp[i][c][0];
      } else
        dp[i + 1][c][0] += dp[i][c][0];
      dp[i + 1][c + 1][1] += dp[i][c][1];
      dp[i + 1][c][1] += dp[i][c][1];
    }
  cout << dp[64][cnt][0] + dp[64][cnt][1] - (cnt == 1) << endl;
}

### 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 dp[65][65][2];
int b[65];
int main() {
  ios::sync_with_stdio(false);
  cout.tie(0);
  cin.tie(0);
  long long n, t, cnt;
  cin >> n >> t;
  if (__builtin_popcountll(t) > 1) {
    cout << 0;
    return 0;
  }
  cnt = __builtin_ctzll(t) + 1;
  cerr << cnt << endl;
  n++;
  for (int i = 0; i < 64; i++) b[i] = (n >> (63 - i)) & 1;
  dp[0][0][0] = 1;
  for (int i = 0; i < 64; i++)
    for (int c = 0; c < 64; c++) {
      if (b[i]) {
        dp[i + 1][c][1] += dp[i][c][0];
        dp[i + 1][c + 1][0] += dp[i][c][0];
      } else
        dp[i + 1][c][0] += dp[i][c][0];
      dp[i + 1][c + 1][1] += dp[i][c][1];
      dp[i + 1][c][1] += dp[i][c][1];
    }
  cout << dp[64][cnt][0] + dp[64][cnt][1] - (cnt == 1) << endl;
}

```