#include <bits/stdc++.h>
using namespace std;
int log2(long long int num) {
  int ans = 0;
  while (num > 1) {
    if (num % 2 != 0) {
      return -1;
    } else {
      ans++;
      num /= 2;
    }
  }
  return ans;
}
struct BinString {
  int len;
  char s[70];
  BinString(long long int num) {
    len = 0;
    while (num > 0) {
      s[len] = ('0' + (num % 2));
      len++;
      num /= 2;
    }
    s[len] = '\0';
  }
  char& operator[](int ind) { return s[len - 1 - ind]; }
};
long long int choose(int n, int k) {
  if (k < 0 || k > n) {
    return 0;
  } else if (k == 0 || k == n) {
    return 1;
  } else {
    return (choose(n - 1, k - 1) * n / k);
  }
}
long long int numOptions(long long int n, int numOnes) {
  BinString b = n;
  long long int ans = 0;
  for (int i = 0; i < b.len; i++) {
    if (b[i] == '1') {
      ans += choose(b.len - 1 - i, numOnes);
      numOnes--;
    }
  }
  if (numOnes == 0) {
    ans++;
  }
  return ans;
}
int main() {
  long long int n, t;
  int logT;
  cin >> n >> t;
  if ((logT = log2(t)) == -1) {
    cout << "0" << endl;
    return 0;
  };
  cout << (numOptions(n + 1, logT + 1) - ((logT == 0) ? 1 : 0)) << "\n";
  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;
int log2(long long int num) {
  int ans = 0;
  while (num > 1) {
    if (num % 2 != 0) {
      return -1;
    } else {
      ans++;
      num /= 2;
    }
  }
  return ans;
}
struct BinString {
  int len;
  char s[70];
  BinString(long long int num) {
    len = 0;
    while (num > 0) {
      s[len] = ('0' + (num % 2));
      len++;
      num /= 2;
    }
    s[len] = '\0';
  }
  char& operator[](int ind) { return s[len - 1 - ind]; }
};
long long int choose(int n, int k) {
  if (k < 0 || k > n) {
    return 0;
  } else if (k == 0 || k == n) {
    return 1;
  } else {
    return (choose(n - 1, k - 1) * n / k);
  }
}
long long int numOptions(long long int n, int numOnes) {
  BinString b = n;
  long long int ans = 0;
  for (int i = 0; i < b.len; i++) {
    if (b[i] == '1') {
      ans += choose(b.len - 1 - i, numOnes);
      numOnes--;
    }
  }
  if (numOnes == 0) {
    ans++;
  }
  return ans;
}
int main() {
  long long int n, t;
  int logT;
  cin >> n >> t;
  if ((logT = log2(t)) == -1) {
    cout << "0" << endl;
    return 0;
  };
  cout << (numOptions(n + 1, logT + 1) - ((logT == 0) ? 1 : 0)) << "\n";
  return 0;
}

```