#include <bits/stdc++.h>
using namespace std;
long long binom[65][65];
long long getbinom(int r, int c) {
  if (c < 0 || c > r || r < 0) return 0;
  if (binom[r][c]) return binom[r][c];
  if (r == 0) return binom[r][c] = 1;
  if (c == 0 || c == r) return binom[r][c] = 1;
  return binom[r][c] = getbinom(r - 1, c) + getbinom(r - 1, c - 1);
}
int main(int argc, char **argv) {
  long long n, t;
  cin >> n >> t;
  n++;
  if ((t & -t) != t) {
    puts("0");
    return 0;
  }
  int p = 0;
  for (p = 0; (1LL << p) != t; ++p)
    ;
  p++;
  long long res = 0;
  int bit = 0;
  for (int i = (63); i >= (0); --i)
    if (n & (1LL << i)) {
      res += getbinom(i, p - bit);
      bit++;
    }
  res += __builtin_popcount(n) == p;
  res -= __builtin_popcount(1) == p;
  cout << res;
  return 0;
}

### Prompt

Your challenge is to write a cpp solution 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>
using namespace std;
long long binom[65][65];
long long getbinom(int r, int c) {
  if (c < 0 || c > r || r < 0) return 0;
  if (binom[r][c]) return binom[r][c];
  if (r == 0) return binom[r][c] = 1;
  if (c == 0 || c == r) return binom[r][c] = 1;
  return binom[r][c] = getbinom(r - 1, c) + getbinom(r - 1, c - 1);
}
int main(int argc, char **argv) {
  long long n, t;
  cin >> n >> t;
  n++;
  if ((t & -t) != t) {
    puts("0");
    return 0;
  }
  int p = 0;
  for (p = 0; (1LL << p) != t; ++p)
    ;
  p++;
  long long res = 0;
  int bit = 0;
  for (int i = (63); i >= (0); --i)
    if (n & (1LL << i)) {
      res += getbinom(i, p - bit);
      bit++;
    }
  res += __builtin_popcount(n) == p;
  res -= __builtin_popcount(1) == p;
  cout << res;
  return 0;
}

```