#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:128777216")
const int null = 0;
using namespace std;
template <class T>
int size(const T &a) {
  return int(a.size());
}
template <class T>
T abs(const T &a) {
  return (a < 0 ? -a : a);
}
int a[100][100];
void init(int m) {
  for (int j = 0; j < m; j++) {
    a[0][j] = 0;
  }
  a[0][m] = 1;
  for (int i = 1; i <= m; i++) {
    for (int j = 0; j <= m; j++) {
      if (j == 0) {
        a[i][j] = a[i - 1][j + 1];
      } else {
        if (j == m) {
          a[i][j] = a[i - 1][j - 1];
        } else {
          a[i][j] = a[i - 1][j - 1] ^ a[i - 1][j + 1];
        }
      }
    }
  }
}
long long cmn[40][40];
void initcmn() {
  for (int i = 0; i < 40; i++) {
    cmn[0][i] = 1;
  }
  for (int i = 1; i < 40; i++) {
    for (int j = 1; j < 40; j++) {
      cmn[i][j] = cmn[i - 1][j - 1] + cmn[i][j - 1];
    }
  }
}
long long calc(long long num, int t) {
  int cur = 0;
  long long res = 0;
  for (int i = 39; i >= 0; i--) {
    if (cur > t) {
      return res;
    }
    if (num & (1LL << i)) {
      res += cmn[t - cur][i];
      cur++;
    }
  }
  if (cur == t) {
    res++;
  }
  return res;
}
int main() {
  initcmn();
  long long n, t;
  cin >> n >> t;
  int lg = 0;
  while (!(t & 1)) {
    t >>= 1;
    lg++;
  }
  if (t != 1) {
    cout << 0 << endl;
    return 0;
  }
  lg++;
  cout << (calc(n + 1, lg) - calc(1, lg)) << endl;
  return 0;
}

### Prompt

Please formulate 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>
#pragma comment(linker, "/STACK:128777216")
const int null = 0;
using namespace std;
template <class T>
int size(const T &a) {
  return int(a.size());
}
template <class T>
T abs(const T &a) {
  return (a < 0 ? -a : a);
}
int a[100][100];
void init(int m) {
  for (int j = 0; j < m; j++) {
    a[0][j] = 0;
  }
  a[0][m] = 1;
  for (int i = 1; i <= m; i++) {
    for (int j = 0; j <= m; j++) {
      if (j == 0) {
        a[i][j] = a[i - 1][j + 1];
      } else {
        if (j == m) {
          a[i][j] = a[i - 1][j - 1];
        } else {
          a[i][j] = a[i - 1][j - 1] ^ a[i - 1][j + 1];
        }
      }
    }
  }
}
long long cmn[40][40];
void initcmn() {
  for (int i = 0; i < 40; i++) {
    cmn[0][i] = 1;
  }
  for (int i = 1; i < 40; i++) {
    for (int j = 1; j < 40; j++) {
      cmn[i][j] = cmn[i - 1][j - 1] + cmn[i][j - 1];
    }
  }
}
long long calc(long long num, int t) {
  int cur = 0;
  long long res = 0;
  for (int i = 39; i >= 0; i--) {
    if (cur > t) {
      return res;
    }
    if (num & (1LL << i)) {
      res += cmn[t - cur][i];
      cur++;
    }
  }
  if (cur == t) {
    res++;
  }
  return res;
}
int main() {
  initcmn();
  long long n, t;
  cin >> n >> t;
  int lg = 0;
  while (!(t & 1)) {
    t >>= 1;
    lg++;
  }
  if (t != 1) {
    cout << 0 << endl;
    return 0;
  }
  lg++;
  cout << (calc(n + 1, lg) - calc(1, lg)) << endl;
  return 0;
}

```