#include <bits/stdc++.h>
using namespace std;
int main() {
  long long M, T;
  cin >> M >> T;
  long long vv = 1, log = 1;
  for (; vv < T; vv <<= 1LL, ++log)
    ;
  if (vv != T) {
    cout << 0 << endl;
    return (0);
  }
  long long dp[61][61], calc = 0, ret = 0LL;
  for (int i = 0; i <= 60; i++) {
    dp[i][0] = 1;
    for (int j = 1; j <= i; j++) {
      dp[i][j] = dp[i - 1][j] + dp[i - 1][j - 1];
    }
  }
  for (int i = 60; i >= 0; i--) {
    if (log < 0) break;
    if (calc + (1LL << i) <= M + 1) {
      ret += dp[i][log--];
      calc += (1LL << i);
    }
  }
  cout << ret + (log == 0) - (T == 1) << endl;
}

### 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;
int main() {
  long long M, T;
  cin >> M >> T;
  long long vv = 1, log = 1;
  for (; vv < T; vv <<= 1LL, ++log)
    ;
  if (vv != T) {
    cout << 0 << endl;
    return (0);
  }
  long long dp[61][61], calc = 0, ret = 0LL;
  for (int i = 0; i <= 60; i++) {
    dp[i][0] = 1;
    for (int j = 1; j <= i; j++) {
      dp[i][j] = dp[i - 1][j] + dp[i - 1][j - 1];
    }
  }
  for (int i = 60; i >= 0; i--) {
    if (log < 0) break;
    if (calc + (1LL << i) <= M + 1) {
      ret += dp[i][log--];
      calc += (1LL << i);
    }
  }
  cout << ret + (log == 0) - (T == 1) << endl;
}

```