#include <bits/stdc++.h>
using namespace std;
const long double pi = 3.14159265359;
template <typename T>
T abs(T x) {
  return x > 0 ? x : -x;
}
template <typename T>
T sqr(T x) {
  return x * x;
}
template <typename T>
void chmin(T &x, T y) {
  x = min(x, y);
}
template <typename T>
void chmax(T &x, T y) {
  x = max(x, y);
}
long long f[60][60];
int main() {
  long long n, t;
  cin >> n >> t;
  f[0][0] = 1;
  for (int i = 1; i < 60; i++) {
    f[i][0] = 1;
    for (int j = 1; j < 60; j++) {
      f[i][j] = f[i - 1][j - 1] + f[i - 1][j];
    }
  }
  long long ans = 0;
  n++;
  if (t == 1) ans--;
  int lv = 0;
  long long len = 1;
  while (n) {
    while (n < len) {
      if (n > len / 2) {
        for (int i = 0; i < 55; i++) {
          if ((1LL << i) == t) {
            ans += f[lv - 1][i];
          }
        }
        if (t & 1) {
          cout << ans << endl;
          return 0;
        }
        t /= 2;
        n -= len / 2;
        lv--;
        len /= 2;
      } else {
        lv--;
        len /= 2;
      }
    }
    assert(len > 0);
    for (int i = 0; i < 55; i++) {
      if ((1LL << i) == t) {
        ans += f[lv][i];
      }
    }
    n -= len;
    lv++;
    len += len;
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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;
const long double pi = 3.14159265359;
template <typename T>
T abs(T x) {
  return x > 0 ? x : -x;
}
template <typename T>
T sqr(T x) {
  return x * x;
}
template <typename T>
void chmin(T &x, T y) {
  x = min(x, y);
}
template <typename T>
void chmax(T &x, T y) {
  x = max(x, y);
}
long long f[60][60];
int main() {
  long long n, t;
  cin >> n >> t;
  f[0][0] = 1;
  for (int i = 1; i < 60; i++) {
    f[i][0] = 1;
    for (int j = 1; j < 60; j++) {
      f[i][j] = f[i - 1][j - 1] + f[i - 1][j];
    }
  }
  long long ans = 0;
  n++;
  if (t == 1) ans--;
  int lv = 0;
  long long len = 1;
  while (n) {
    while (n < len) {
      if (n > len / 2) {
        for (int i = 0; i < 55; i++) {
          if ((1LL << i) == t) {
            ans += f[lv - 1][i];
          }
        }
        if (t & 1) {
          cout << ans << endl;
          return 0;
        }
        t /= 2;
        n -= len / 2;
        lv--;
        len /= 2;
      } else {
        lv--;
        len /= 2;
      }
    }
    assert(len > 0);
    for (int i = 0; i < 55; i++) {
      if ((1LL << i) == t) {
        ans += f[lv][i];
      }
    }
    n -= len;
    lv++;
    len += len;
  }
  cout << ans << endl;
  return 0;
}

```