#include <bits/stdc++.h>
long long C[60][60];
void calC() {
  for (int i = 0; i < 60; i++) C[i][0] = C[i][i] = 1;
  for (int i = 1; i < 60; i++)
    for (int j = 1; j < i; j++) C[i][j] = C[i - 1][j] + C[i - 1][j - 1];
}
int main() {
  calC();
  long long n, t;
  scanf("%I64d%I64d", &n, &t);
  n++;
  if (t & (t - 1)) {
    printf("0\n");
    return 0;
  }
  long long p = 1 + log2(t);
  long long ans = 0, res = p;
  for (int i = 60 - 1; i >= 0; i--)
    if (n & (1LL << i)) {
      ans += C[i][res];
      if (res == p && res == 1) ans--;
      res--;
      if (res < 0) break;
    }
  if (__builtin_popcount(n) == p) ans++;
  printf("%I64d\n", ans);
  return 0;
}

### Prompt

Please create a solution in CPP 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>
long long C[60][60];
void calC() {
  for (int i = 0; i < 60; i++) C[i][0] = C[i][i] = 1;
  for (int i = 1; i < 60; i++)
    for (int j = 1; j < i; j++) C[i][j] = C[i - 1][j] + C[i - 1][j - 1];
}
int main() {
  calC();
  long long n, t;
  scanf("%I64d%I64d", &n, &t);
  n++;
  if (t & (t - 1)) {
    printf("0\n");
    return 0;
  }
  long long p = 1 + log2(t);
  long long ans = 0, res = p;
  for (int i = 60 - 1; i >= 0; i--)
    if (n & (1LL << i)) {
      ans += C[i][res];
      if (res == p && res == 1) ans--;
      res--;
      if (res < 0) break;
    }
  if (__builtin_popcount(n) == p) ans++;
  printf("%I64d\n", ans);
  return 0;
}

```