#include <bits/stdc++.h>
using namespace std;
long long N, M;
int main() {
  std::ios_base::sync_with_stdio(false);
  cin >> N >> M;
  int n = 0;
  while ((1LL << n) < M) {
    ++n;
  }
  if ((1LL << n) != M) {
    printf("0\n");
    return 0;
  }
  ++n;
  int dig[100], ds = 0;
  for (long long mt = N + 1; mt > 0; mt >>= 1) {
    dig[ds++] = mt & 1;
  }
  reverse(dig, dig + ds);
  long long dp[2][2][100];
  int cur = 0, pre = 1;
  fill(dp[cur][0], dp[cur][2], 0LL);
  dp[cur][0][0] = 1;
  for (int i = 0; i < ds; ++i) {
    swap(cur, pre);
    fill(dp[cur][0], dp[cur][2], 0LL);
    for (int o = 0; o < 2; ++o) {
      for (int j = 0; j < 100; ++j) {
        if (dp[pre][o][j]) {
          if (o == 0) {
            if (dig[i] == 1) {
              dp[cur][1][j] += dp[pre][o][j];
              dp[cur][0][j + 1] += dp[pre][o][j];
            } else {
              dp[cur][0][j] += dp[pre][o][j];
            }
          } else if (o == 1) {
            dp[cur][1][j] += dp[pre][o][j];
            dp[cur][1][j + 1] += dp[pre][o][j];
          }
        }
      }
    }
  }
  long long ans = dp[cur][0][n] + dp[cur][1][n];
  if (M == 1) {
    ans -= 1LL;
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Please create a solution in cpp 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 N, M;
int main() {
  std::ios_base::sync_with_stdio(false);
  cin >> N >> M;
  int n = 0;
  while ((1LL << n) < M) {
    ++n;
  }
  if ((1LL << n) != M) {
    printf("0\n");
    return 0;
  }
  ++n;
  int dig[100], ds = 0;
  for (long long mt = N + 1; mt > 0; mt >>= 1) {
    dig[ds++] = mt & 1;
  }
  reverse(dig, dig + ds);
  long long dp[2][2][100];
  int cur = 0, pre = 1;
  fill(dp[cur][0], dp[cur][2], 0LL);
  dp[cur][0][0] = 1;
  for (int i = 0; i < ds; ++i) {
    swap(cur, pre);
    fill(dp[cur][0], dp[cur][2], 0LL);
    for (int o = 0; o < 2; ++o) {
      for (int j = 0; j < 100; ++j) {
        if (dp[pre][o][j]) {
          if (o == 0) {
            if (dig[i] == 1) {
              dp[cur][1][j] += dp[pre][o][j];
              dp[cur][0][j + 1] += dp[pre][o][j];
            } else {
              dp[cur][0][j] += dp[pre][o][j];
            }
          } else if (o == 1) {
            dp[cur][1][j] += dp[pre][o][j];
            dp[cur][1][j + 1] += dp[pre][o][j];
          }
        }
      }
    }
  }
  long long ans = dp[cur][0][n] + dp[cur][1][n];
  if (M == 1) {
    ans -= 1LL;
  }
  cout << ans << endl;
  return 0;
}

```