#include <bits/stdc++.h>
using namespace std;
long long n, t;
long long ans;
long long DP[51][51];
long long pow2[51];
long long comp(int n, int k, long long rows) {
  if (rows <= 0) return 0;
  if (k == 0) return 1;
  if (n == 0) return 0;
  if (rows >= pow2[n]) {
    if (DP[n][k] == -1)
      DP[n][k] = comp(n - 1, k, rows) + comp(n - 1, k - 1, rows);
    return DP[n][k];
  }
  long long res = 0;
  res = comp(n - 1, k, rows);
  res += comp(n - 1, k - 1, rows - pow2[n - 1]);
  return res;
}
int main() {
  int i, j, k;
  for (i = 0; i < 50; i++)
    for (j = 0; j < 50; j++) DP[i][j] = -1;
  pow2[0] = 1;
  for (i = 1; i <= 50; i++) pow2[i] = pow2[i - 1] * 2;
  scanf("%I64d%I64d", &n, &t);
  n++;
  j = 0;
  k = 0;
  while (t > 1) {
    if (t % 2 == 1) {
      printf("0\n");
      return 0;
    }
    t /= 2;
    k++;
  }
  long long ans = 0;
  while (n > 0) {
    ans += comp(j, k, n);
    n -= pow2[j];
    j++;
  }
  if (k == 0) ans--;
  printf("%I64d\n", ans);
  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;
long long n, t;
long long ans;
long long DP[51][51];
long long pow2[51];
long long comp(int n, int k, long long rows) {
  if (rows <= 0) return 0;
  if (k == 0) return 1;
  if (n == 0) return 0;
  if (rows >= pow2[n]) {
    if (DP[n][k] == -1)
      DP[n][k] = comp(n - 1, k, rows) + comp(n - 1, k - 1, rows);
    return DP[n][k];
  }
  long long res = 0;
  res = comp(n - 1, k, rows);
  res += comp(n - 1, k - 1, rows - pow2[n - 1]);
  return res;
}
int main() {
  int i, j, k;
  for (i = 0; i < 50; i++)
    for (j = 0; j < 50; j++) DP[i][j] = -1;
  pow2[0] = 1;
  for (i = 1; i <= 50; i++) pow2[i] = pow2[i - 1] * 2;
  scanf("%I64d%I64d", &n, &t);
  n++;
  j = 0;
  k = 0;
  while (t > 1) {
    if (t % 2 == 1) {
      printf("0\n");
      return 0;
    }
    t /= 2;
    k++;
  }
  long long ans = 0;
  while (n > 0) {
    ans += comp(j, k, n);
    n -= pow2[j];
    j++;
  }
  if (k == 0) ans--;
  printf("%I64d\n", ans);
  return 0;
}

```