#include <bits/stdc++.h>
using namespace std;
const int logn = 44;
long long n, t, dp[logn + 1][logn + 1][2];
int bits = 0;
inline int bit(long long x, int p) { return (x >> p) & 1; }
int main() {
  cin >> n >> t, n++;
  for (; t != 1; t >>= 1) {
    if (t & 1) cout << 0 << endl, exit(0);
    bits++;
  }
  bits++;
  dp[0][0][1] = 1;
  for (int i = 0; i < logn; i++)
    for (int j = 0; j <= i; j++)
      for (int z = 0; z < 2; z++) {
        if (!dp[i][j][z]) continue;
        int b = bit(n, logn - 1 - i);
        for (int t = 0; t < 2; t++) {
          if (z && t > b) continue;
          int nz = (z && t == b) ? 1 : 0;
          dp[i + 1][j + t][nz] += dp[i][j][z];
        }
      }
  cout << dp[logn][bits][0] + dp[logn][bits][1] - (bits == 1 ? 1 : 0) << endl;
  return 0;
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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 logn = 44;
long long n, t, dp[logn + 1][logn + 1][2];
int bits = 0;
inline int bit(long long x, int p) { return (x >> p) & 1; }
int main() {
  cin >> n >> t, n++;
  for (; t != 1; t >>= 1) {
    if (t & 1) cout << 0 << endl, exit(0);
    bits++;
  }
  bits++;
  dp[0][0][1] = 1;
  for (int i = 0; i < logn; i++)
    for (int j = 0; j <= i; j++)
      for (int z = 0; z < 2; z++) {
        if (!dp[i][j][z]) continue;
        int b = bit(n, logn - 1 - i);
        for (int t = 0; t < 2; t++) {
          if (z && t > b) continue;
          int nz = (z && t == b) ? 1 : 0;
          dp[i + 1][j + t][nz] += dp[i][j][z];
        }
      }
  cout << dp[logn][bits][0] + dp[logn][bits][1] - (bits == 1 ? 1 : 0) << endl;
  return 0;
}

```