#include <bits/stdc++.h>
using namespace std;
int ord2(long long k) {
  if (k == 0) return 0;
  int result = 0;
  while (k != 1) {
    if (k % 2 == 1) return -1;
    k /= 2;
    result++;
  }
  return result;
}
long long C(int n, int k) {
  if (k > n) return 0;
  long long result = 1;
  for (int i = 1; i <= k; i++) {
    result *= (n + 1 - i);
    result /= i;
  }
  return result;
}
long long count(long long n, int k) {
  long long result = 0;
  for (int bit = 55; bit >= 0; bit--) {
    if (k <= -1) break;
    long long pow2 = 1LL << bit;
    if (pow2 > n)
      continue;
    else {
      long long temp = C(bit, k);
      result += temp;
      n -= pow2;
      k--;
    }
  }
  if (k == 0) result++;
  return result;
}
long long solve(long long n, long long t) {
  int ord = ord2(t);
  if (ord == -1) return 0;
  int target = ord + 1;
  long long result = count(n + 1, target) - count(1, target);
  return result;
}
int main() {
  long long n, t;
  cin >> n >> t;
  cout << solve(n, t);
  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>
using namespace std;
int ord2(long long k) {
  if (k == 0) return 0;
  int result = 0;
  while (k != 1) {
    if (k % 2 == 1) return -1;
    k /= 2;
    result++;
  }
  return result;
}
long long C(int n, int k) {
  if (k > n) return 0;
  long long result = 1;
  for (int i = 1; i <= k; i++) {
    result *= (n + 1 - i);
    result /= i;
  }
  return result;
}
long long count(long long n, int k) {
  long long result = 0;
  for (int bit = 55; bit >= 0; bit--) {
    if (k <= -1) break;
    long long pow2 = 1LL << bit;
    if (pow2 > n)
      continue;
    else {
      long long temp = C(bit, k);
      result += temp;
      n -= pow2;
      k--;
    }
  }
  if (k == 0) result++;
  return result;
}
long long solve(long long n, long long t) {
  int ord = ord2(t);
  if (ord == -1) return 0;
  int target = ord + 1;
  long long result = count(n + 1, target) - count(1, target);
  return result;
}
int main() {
  long long n, t;
  cin >> n >> t;
  cout << solve(n, t);
  return 0;
}

```