#include <bits/stdc++.h>
using namespace std;
long long n, t;
long long num[50][50];
int main() {
  while (scanf("%I64d%I64d", &n, &t) == 2) {
    int tot = 0;
    while (t % 2 == 0) {
      t = t / 2;
      tot++;
    }
    if (t != 1) {
      printf("0\n");
      continue;
    }
    memset(num, 0, sizeof(num));
    for (int i = 0; i <= 45; i++) num[i][0] = 1;
    for (int i = 1; i <= 45; i++)
      for (int j = 1; j <= 45; j++)
        num[i][j] = num[i - 1][j] + num[i - 1][j - 1];
    long long tp = 0;
    int jc = 1;
    long long ni = 2;
    long long ans = 0;
    for (;; ni = ni * 2, jc++)
      if (tp + ni <= n) {
        tp = tp + ni;
        ans = ans + num[jc][tot];
      } else
        break;
    n = n - tp;
    ni = ni / 2;
    jc = jc - 1;
    for (; jc >= 0; jc--, ni = ni / 2) {
      if (tot < 0) break;
      if (n >= ni) {
        n = n - ni;
        ans = ans + num[jc][tot];
        tot--;
      }
    }
    printf("%I64d\n", ans);
  }
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 n, t;
long long num[50][50];
int main() {
  while (scanf("%I64d%I64d", &n, &t) == 2) {
    int tot = 0;
    while (t % 2 == 0) {
      t = t / 2;
      tot++;
    }
    if (t != 1) {
      printf("0\n");
      continue;
    }
    memset(num, 0, sizeof(num));
    for (int i = 0; i <= 45; i++) num[i][0] = 1;
    for (int i = 1; i <= 45; i++)
      for (int j = 1; j <= 45; j++)
        num[i][j] = num[i - 1][j] + num[i - 1][j - 1];
    long long tp = 0;
    int jc = 1;
    long long ni = 2;
    long long ans = 0;
    for (;; ni = ni * 2, jc++)
      if (tp + ni <= n) {
        tp = tp + ni;
        ans = ans + num[jc][tot];
      } else
        break;
    n = n - tp;
    ni = ni / 2;
    jc = jc - 1;
    for (; jc >= 0; jc--, ni = ni / 2) {
      if (tot < 0) break;
      if (n >= ni) {
        n = n - ni;
        ans = ans + num[jc][tot];
        tot--;
      }
    }
    printf("%I64d\n", ans);
  }
  return 0;
}

```