#include <bits/stdc++.h>
using namespace std;
long long n, t;
long long dp[55][55];
long long solve(long long n, int p) {
  if (p == 0) return 1;
  for (int i = 50; i >= 0; --i) {
    if ((1ll << i) & n) {
      return dp[i][p] + solve(n - (1ll << i), p - 1);
    }
  }
  return 0;
}
int main(void) {
  dp[0][0] = 1;
  for (int i = 0; i < 50; ++i) {
    for (int j = 0; j < 50; ++j) {
      dp[i + 1][j] = dp[i][j] + (j >= 1 ? dp[i][j - 1] : 0);
    }
  }
  cin >> n >> t;
  ++n;
  int p = 0;
  while (1ll << p < t) ++p;
  if (1ll << p != t) {
    printf("0\n");
    return 0;
  }
  ++p;
  cout << solve(n, p) - (t == 1 ? 1 : 0) << endl;
  return 0;
}

### Prompt

In cpp, your task is to solve 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 n, t;
long long dp[55][55];
long long solve(long long n, int p) {
  if (p == 0) return 1;
  for (int i = 50; i >= 0; --i) {
    if ((1ll << i) & n) {
      return dp[i][p] + solve(n - (1ll << i), p - 1);
    }
  }
  return 0;
}
int main(void) {
  dp[0][0] = 1;
  for (int i = 0; i < 50; ++i) {
    for (int j = 0; j < 50; ++j) {
      dp[i + 1][j] = dp[i][j] + (j >= 1 ? dp[i][j - 1] : 0);
    }
  }
  cin >> n >> t;
  ++n;
  int p = 0;
  while (1ll << p < t) ++p;
  if (1ll << p != t) {
    printf("0\n");
    return 0;
  }
  ++p;
  cout << solve(n, p) - (t == 1 ? 1 : 0) << endl;
  return 0;
}

```