#include <bits/stdc++.h>
using namespace std;
int m;
long long n, t;
long long val[60][60];
int main() {
  val[0][0] = 1;
  for (int i = 1; i <= 50; i++) {
    for (int j = 0; j < i; j++) {
      val[i][j] = val[i - 1][j];
    }
    for (int j = 1; j <= i; j++) val[i][j] += val[i - 1][j - 1];
  }
  cin >> n >> t;
  int idx = 0;
  while (t != 1) {
    if (t & 1) {
      cout << 0 << endl;
      return 0;
    }
    idx++;
    t >>= 1;
  }
  long long res = 0;
  for (long long i = 1; n >= (1LL << i); i++) {
    res += val[i][idx];
    n -= (1LL << i);
  }
  for (long long i = 40; i >= 0; i--) {
    if ((1LL << i) <= n) {
      if (idx >= 0) {
        res += val[i][idx];
        idx--;
      }
      n -= (1LL << i);
    }
  }
  cout << res << endl;
  return 0;
}

### Prompt

Your task is to create 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 m;
long long n, t;
long long val[60][60];
int main() {
  val[0][0] = 1;
  for (int i = 1; i <= 50; i++) {
    for (int j = 0; j < i; j++) {
      val[i][j] = val[i - 1][j];
    }
    for (int j = 1; j <= i; j++) val[i][j] += val[i - 1][j - 1];
  }
  cin >> n >> t;
  int idx = 0;
  while (t != 1) {
    if (t & 1) {
      cout << 0 << endl;
      return 0;
    }
    idx++;
    t >>= 1;
  }
  long long res = 0;
  for (long long i = 1; n >= (1LL << i); i++) {
    res += val[i][idx];
    n -= (1LL << i);
  }
  for (long long i = 40; i >= 0; i--) {
    if ((1LL << i) <= n) {
      if (idx >= 0) {
        res += val[i][idx];
        idx--;
      }
      n -= (1LL << i);
    }
  }
  cout << res << endl;
  return 0;
}

```