#include <bits/stdc++.h>
long long C[100][100];
long long ans = 0;
int main() {
  long long n, t;
  std::cin >> n >> t;
  n++;
  if (t & (t - 1)) return std::cout << 0 << std::endl, 0;
  int d = 0;
  if (t == 1) ans--;
  while (t) d++, t = t >> 1;
  C[0][0] = 1;
  for (int i = 1; i <= 62; i++) C[i][0] = 1;
  for (int i = 1; i <= 62; i++)
    for (int j = 1; j <= i; j++) C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
  for (int i = 62; i >= 0; i--)
    if ((n >> i) & 1)
      if (d >= 0) ans += C[i][d--];
  if (d == 0) ans++;
  std::cout << ans << std::endl;
}

### 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>
long long C[100][100];
long long ans = 0;
int main() {
  long long n, t;
  std::cin >> n >> t;
  n++;
  if (t & (t - 1)) return std::cout << 0 << std::endl, 0;
  int d = 0;
  if (t == 1) ans--;
  while (t) d++, t = t >> 1;
  C[0][0] = 1;
  for (int i = 1; i <= 62; i++) C[i][0] = 1;
  for (int i = 1; i <= 62; i++)
    for (int j = 1; j <= i; j++) C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
  for (int i = 62; i >= 0; i--)
    if ((n >> i) & 1)
      if (d >= 0) ans += C[i][d--];
  if (d == 0) ans++;
  std::cout << ans << std::endl;
}

```