#include <bits/stdc++.h>
using namespace std;
const int N = 100000 + 111;
long long c[111][111];
void init() {
  int i, j;
  c[0][0] = 1;
  for (i = 1; i < 100; i++) {
    c[i][0] = c[i - 1][0];
    for (j = 1; j <= i; j++) {
      c[i][j] = c[i - 1][j - 1] + c[i - 1][j];
    }
  }
}
int main() {
  int i, j;
  long long n, m;
  init();
  cin >> n >> m;
  if (m & (m - 1)) {
    puts("0");
    return 0;
  }
  long long d = 0, tmp = m;
  while (tmp) {
    d++;
    tmp >>= 1;
  }
  n++;
  n++;
  long long ans = d == 1 ? -1 : 0;
  for (i = 60; i >= 0; i--) {
    long long tp = 1ll << i;
    if (n < tp) continue;
    if (n & tp) {
      if (d < 0) break;
      ans += c[i][d];
      d--;
      n -= (1ll << i);
    }
  }
  cout << ans << endl;
  return 0;
}

### Prompt

Develop a solution in cpp 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 int N = 100000 + 111;
long long c[111][111];
void init() {
  int i, j;
  c[0][0] = 1;
  for (i = 1; i < 100; i++) {
    c[i][0] = c[i - 1][0];
    for (j = 1; j <= i; j++) {
      c[i][j] = c[i - 1][j - 1] + c[i - 1][j];
    }
  }
}
int main() {
  int i, j;
  long long n, m;
  init();
  cin >> n >> m;
  if (m & (m - 1)) {
    puts("0");
    return 0;
  }
  long long d = 0, tmp = m;
  while (tmp) {
    d++;
    tmp >>= 1;
  }
  n++;
  n++;
  long long ans = d == 1 ? -1 : 0;
  for (i = 60; i >= 0; i--) {
    long long tp = 1ll << i;
    if (n < tp) continue;
    if (n & tp) {
      if (d < 0) break;
      ans += c[i][d];
      d--;
      n -= (1ll << i);
    }
  }
  cout << ans << endl;
  return 0;
}

```