#include <bits/stdc++.h>
using namespace std;
const int LG = 50;
long long pre[LG][LG];
int main() {
  pre[0][0] = 1;
  for (int i = 1; i < LG; i++) {
    for (int j = 0; j < LG; j++) {
      pre[i][j] = pre[i - 1][j];
      if (j) pre[i][j] += pre[i - 1][j - 1];
    }
  }
  long long n, t;
  int lg = 0;
  cin >> n >> t;
  while (!(t & 1)) t >>= 1, lg++;
  if (t > 1) return cout << 0 << endl, 0;
  long long ans = 0;
  for (int i = 1; (1LL << i) <= n; i++) {
    ans += pre[i][lg];
    n -= 1LL << i;
  }
  int coff = 0;
  for (int i = LG - 1; ~i; i--)
    if ((1LL << i) <= n) {
      if (lg >= coff) ans += pre[i][lg - coff];
      coff++;
      n -= 1LL << i;
    }
  cout << ans << 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;
const int LG = 50;
long long pre[LG][LG];
int main() {
  pre[0][0] = 1;
  for (int i = 1; i < LG; i++) {
    for (int j = 0; j < LG; j++) {
      pre[i][j] = pre[i - 1][j];
      if (j) pre[i][j] += pre[i - 1][j - 1];
    }
  }
  long long n, t;
  int lg = 0;
  cin >> n >> t;
  while (!(t & 1)) t >>= 1, lg++;
  if (t > 1) return cout << 0 << endl, 0;
  long long ans = 0;
  for (int i = 1; (1LL << i) <= n; i++) {
    ans += pre[i][lg];
    n -= 1LL << i;
  }
  int coff = 0;
  for (int i = LG - 1; ~i; i--)
    if ((1LL << i) <= n) {
      if (lg >= coff) ans += pre[i][lg - coff];
      coff++;
      n -= 1LL << i;
    }
  cout << ans << endl;
  return 0;
}

```