#include <bits/stdc++.h>
using namespace std;
long long a, b;
int p;
int tot, len;
int c[60];
long long dp[60][60][2];
long long calc(long long n, long long t) {
  if ((t & (-t)) != t) return 0;
  for (; t; t >>= 1) tot++;
  n++;
  for (; n;) {
    c[++len] = n & 1;
    n >>= 1;
  }
  for (int i = 1, j = len; i < j; i++, j--) swap(c[i], c[j]);
  dp[0][0][1] = 1;
  for (int i = 0; i < len; i++)
    for (int j = 0; j <= tot; j++) {
      dp[i + 1][j + 1][0] += dp[i][j][0];
      dp[i + 1][j][0] += dp[i][j][0];
      if (c[i + 1] == 1) {
        dp[i + 1][j + 1][1] += dp[i][j][1];
        dp[i + 1][j][0] += dp[i][j][1];
      } else
        dp[i + 1][j][1] += dp[i][j][1];
    }
  return dp[len][tot][0] + dp[len][tot][1] - (tot == 1);
}
int main() {
  cin >> a >> b;
  cout << calc(a, b) << 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;
long long a, b;
int p;
int tot, len;
int c[60];
long long dp[60][60][2];
long long calc(long long n, long long t) {
  if ((t & (-t)) != t) return 0;
  for (; t; t >>= 1) tot++;
  n++;
  for (; n;) {
    c[++len] = n & 1;
    n >>= 1;
  }
  for (int i = 1, j = len; i < j; i++, j--) swap(c[i], c[j]);
  dp[0][0][1] = 1;
  for (int i = 0; i < len; i++)
    for (int j = 0; j <= tot; j++) {
      dp[i + 1][j + 1][0] += dp[i][j][0];
      dp[i + 1][j][0] += dp[i][j][0];
      if (c[i + 1] == 1) {
        dp[i + 1][j + 1][1] += dp[i][j][1];
        dp[i + 1][j][0] += dp[i][j][1];
      } else
        dp[i + 1][j][1] += dp[i][j][1];
    }
  return dp[len][tot][0] + dp[len][tot][1] - (tot == 1);
}
int main() {
  cin >> a >> b;
  cout << calc(a, b) << endl;
}

```