#include <bits/stdc++.h>
using namespace std;
long long C(int n, int a) {
  long long ans = 1;
  if (n > 2 * a) a = n - a;
  for (int i = 1; i <= a; i++) {
    ans *= n - i + 1;
    ans /= i;
  }
  return ans;
}
int main() {
  long long n, t;
  cin >> n >> t;
  int a = 0;
  while (t > 1) {
    if (t % 2) {
      cout << 0;
      return 0;
    }
    a++;
    t /= 2;
  }
  a += 1;
  n += 2;
  vector<int> digit;
  while (n > 0) {
    digit.push_back(n % 2);
    n /= 2;
  }
  long long ans = 0;
  if (a == 1) ans -= 1;
  for (int i = digit.size() - 1; i >= 0; i--) {
    cerr << ans << endl;
    if (digit[i]) {
      ans += C(i, a);
      a--;
    }
  }
  cout << ans;
  return 0;
}

### Prompt

Generate 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 C(int n, int a) {
  long long ans = 1;
  if (n > 2 * a) a = n - a;
  for (int i = 1; i <= a; i++) {
    ans *= n - i + 1;
    ans /= i;
  }
  return ans;
}
int main() {
  long long n, t;
  cin >> n >> t;
  int a = 0;
  while (t > 1) {
    if (t % 2) {
      cout << 0;
      return 0;
    }
    a++;
    t /= 2;
  }
  a += 1;
  n += 2;
  vector<int> digit;
  while (n > 0) {
    digit.push_back(n % 2);
    n /= 2;
  }
  long long ans = 0;
  if (a == 1) ans -= 1;
  for (int i = digit.size() - 1; i >= 0; i--) {
    cerr << ans << endl;
    if (digit[i]) {
      ans += C(i, a);
      a--;
    }
  }
  cout << ans;
  return 0;
}

```