#include <bits/stdc++.h>
using namespace std;
long long int ncr(long long int n, long long int r) {
  if (r > n) return 0;
  if (n == r) return 1;
  if (r == 0) return 1;
  if (r == 1) return n;
  long long int ans = 1;
  for (long long int i = 1; i <= r; i++) {
    ans *= (n - i + 1);
    ans /= i;
  }
  return ans;
}
long long int solve(long long int n, long long int k) {
  if (k == 0) return 1;
  if (n == 1) {
    if (k >= 2) return 0;
    return k;
  }
  if (n == 0) return 0;
  long long int temp = 1, cnt = 0;
  while (2 * temp <= n) {
    cnt++;
    temp *= 2;
  }
  return ncr(cnt, k) + solve(n - temp, k - 1);
}
int main() {
  long long int i, j, n, t, cnt = 0, ans;
  cin >> n >> t;
  while (t % 2 == 0) {
    cnt++;
    t = t / 2;
  }
  if (t != 1) {
    cout << "0";
    return 0;
  }
  cnt++;
  ans = solve(n + 1, cnt);
  if (cnt == 1) ans--;
  cout << ans;
  return 0;
}

### Prompt

In CPP, your task is to solve 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 int ncr(long long int n, long long int r) {
  if (r > n) return 0;
  if (n == r) return 1;
  if (r == 0) return 1;
  if (r == 1) return n;
  long long int ans = 1;
  for (long long int i = 1; i <= r; i++) {
    ans *= (n - i + 1);
    ans /= i;
  }
  return ans;
}
long long int solve(long long int n, long long int k) {
  if (k == 0) return 1;
  if (n == 1) {
    if (k >= 2) return 0;
    return k;
  }
  if (n == 0) return 0;
  long long int temp = 1, cnt = 0;
  while (2 * temp <= n) {
    cnt++;
    temp *= 2;
  }
  return ncr(cnt, k) + solve(n - temp, k - 1);
}
int main() {
  long long int i, j, n, t, cnt = 0, ans;
  cin >> n >> t;
  while (t % 2 == 0) {
    cnt++;
    t = t / 2;
  }
  if (t != 1) {
    cout << "0";
    return 0;
  }
  cnt++;
  ans = solve(n + 1, cnt);
  if (cnt == 1) ans--;
  cout << ans;
  return 0;
}

```