#include <bits/stdc++.h>
using namespace std;
int naive(int M) {
  bool val[2010][2010] = {};
  val[0][M] = true;
  for (int i = 1; i <= M; i++) {
    for (int j = 0; j <= M; j++) {
      if (0 < j) val[i][j] ^= val[i - 1][j - 1];
      if (j + 1 <= M) val[i][j] ^= val[i - 1][j + 1];
    }
  }
  int res = 0;
  for (int i = 0; i <= M; i++) {
    res += val[M][i];
  }
  return res;
}
int main() {
  long long N, T;
  scanf("%lld%lld", &N, &T);
  int counter = 1;
  while (T > 1) {
    if (T % 2 != 0) {
      puts("0");
      return 0;
    }
    counter++;
    T /= 2;
  }
  long long dp[50][50][2] = {};
  int B = 44;
  dp[0][0][0] = 1;
  N++;
  for (int i = 0; i <= B; i++) {
    int p = !!(N & (1LL << (B - i)));
    for (int j = 0; j <= i; j++) {
      if (p) {
        dp[i + 1][j][1] += dp[i][j][0] + dp[i][j][1];
        dp[i + 1][j + 1][0] += dp[i][j][0];
        dp[i + 1][j + 1][1] += dp[i][j][1];
      } else {
        dp[i + 1][j][0] += dp[i][j][0];
        dp[i + 1][j][1] += dp[i][j][1];
        dp[i + 1][j + 1][1] += dp[i][j][1];
      }
    }
  }
  long long ans = dp[B + 1][counter][0] + dp[B + 1][counter][1];
  if (counter == 1) ans--;
  cout << ans << endl;
  return 0;
}

### Prompt

Generate 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 naive(int M) {
  bool val[2010][2010] = {};
  val[0][M] = true;
  for (int i = 1; i <= M; i++) {
    for (int j = 0; j <= M; j++) {
      if (0 < j) val[i][j] ^= val[i - 1][j - 1];
      if (j + 1 <= M) val[i][j] ^= val[i - 1][j + 1];
    }
  }
  int res = 0;
  for (int i = 0; i <= M; i++) {
    res += val[M][i];
  }
  return res;
}
int main() {
  long long N, T;
  scanf("%lld%lld", &N, &T);
  int counter = 1;
  while (T > 1) {
    if (T % 2 != 0) {
      puts("0");
      return 0;
    }
    counter++;
    T /= 2;
  }
  long long dp[50][50][2] = {};
  int B = 44;
  dp[0][0][0] = 1;
  N++;
  for (int i = 0; i <= B; i++) {
    int p = !!(N & (1LL << (B - i)));
    for (int j = 0; j <= i; j++) {
      if (p) {
        dp[i + 1][j][1] += dp[i][j][0] + dp[i][j][1];
        dp[i + 1][j + 1][0] += dp[i][j][0];
        dp[i + 1][j + 1][1] += dp[i][j][1];
      } else {
        dp[i + 1][j][0] += dp[i][j][0];
        dp[i + 1][j][1] += dp[i][j][1];
        dp[i + 1][j + 1][1] += dp[i][j][1];
      }
    }
  }
  long long ans = dp[B + 1][counter][0] + dp[B + 1][counter][1];
  if (counter == 1) ans--;
  cout << ans << endl;
  return 0;
}

```