#include <bits/stdc++.h>
using namespace std;
long long dp[105][105];
long long N, T;
long long init(long long n, long long r) {
  if (n < r) {
    if (n < 0) {
      return 0;
    }
    if (r < 0) {
      return 0;
    }
    return dp[n][r] = 0;
  }
  if (r == 0) {
    return dp[n][r] = 1;
  }
  if (r == 1) {
    return dp[n][r] = n;
  }
  if (n == r) {
    return dp[n][r] = 1;
  }
  if (dp[n][r] > 0) {
    return dp[n][r];
  }
  return dp[n][r] = (init(n - 1, r) + init(n - 1, r - 1));
}
bool NotPower(long long x) {
  int Count = 0;
  int temp = x;
  while (temp) {
    temp >>= 1;
    Count++;
  }
  long long y = 1;
  temp = Count;
  while (Count) {
    y *= 2ll;
    Count--;
  }
  return (x == y);
}
int main() {
  cin >> N >> T;
  ++N;
  if (T & (T - 1)) {
    cout << 0 << endl;
  } else {
    int n = 0;
    long long Ans = 0;
    if (T == 1) {
      Ans--;
    }
    while (T) {
      T >>= 1;
      n++;
    }
    for (long long i = 62; i >= 0; --i) {
      if ((N >> i) & 1) {
        if (n >= 0) {
          Ans += init(i, n);
          --n;
        }
      }
    }
    if (n == 0) {
      Ans++;
    }
    cout << Ans << 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 dp[105][105];
long long N, T;
long long init(long long n, long long r) {
  if (n < r) {
    if (n < 0) {
      return 0;
    }
    if (r < 0) {
      return 0;
    }
    return dp[n][r] = 0;
  }
  if (r == 0) {
    return dp[n][r] = 1;
  }
  if (r == 1) {
    return dp[n][r] = n;
  }
  if (n == r) {
    return dp[n][r] = 1;
  }
  if (dp[n][r] > 0) {
    return dp[n][r];
  }
  return dp[n][r] = (init(n - 1, r) + init(n - 1, r - 1));
}
bool NotPower(long long x) {
  int Count = 0;
  int temp = x;
  while (temp) {
    temp >>= 1;
    Count++;
  }
  long long y = 1;
  temp = Count;
  while (Count) {
    y *= 2ll;
    Count--;
  }
  return (x == y);
}
int main() {
  cin >> N >> T;
  ++N;
  if (T & (T - 1)) {
    cout << 0 << endl;
  } else {
    int n = 0;
    long long Ans = 0;
    if (T == 1) {
      Ans--;
    }
    while (T) {
      T >>= 1;
      n++;
    }
    for (long long i = 62; i >= 0; --i) {
      if ((N >> i) & 1) {
        if (n >= 0) {
          Ans += init(i, n);
          --n;
        }
      }
    }
    if (n == 0) {
      Ans++;
    }
    cout << Ans << endl;
  }
  return 0;
}

```