#include <bits/stdc++.h>
using namespace std;
long long com[64][64];
long long gc(int n, int k) {
  if (n < k) return 0;
  if (n == k) return 1;
  if (k == 0) return 1;
  if (com[n][k] != -1) return com[n][k];
  com[n][k] = gc(n - 1, k) + gc(n - 1, k - 1);
  return com[n][k];
}
long long n, t;
int k;
int getk(long long x) {
  int res = 0;
  while (x > 1) {
    if (x & 1) return -1;
    res++;
    x = x / 2;
  }
  return res;
}
int main() {
  int i, j;
  long long ans = 0;
  memset(com, -1, sizeof(com));
  cin >> n >> t;
  k = getk(t);
  if (k == -1) {
    printf("0\n");
    return 0;
  }
  k++;
  long long head = 0;
  int rest = k;
  for (i = 60; i >= 0; i--) {
    if (head + (1LL << i) <= n + 1) {
      head += (1LL << i);
      if (rest > 1)
        ans = ans + gc(i, rest);
      else
        ans = ans + gc(i + 1, rest);
      rest--;
    }
    if (rest == 0) break;
  }
  if (t == 1) ans--;
  cout << ans << endl;
  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 com[64][64];
long long gc(int n, int k) {
  if (n < k) return 0;
  if (n == k) return 1;
  if (k == 0) return 1;
  if (com[n][k] != -1) return com[n][k];
  com[n][k] = gc(n - 1, k) + gc(n - 1, k - 1);
  return com[n][k];
}
long long n, t;
int k;
int getk(long long x) {
  int res = 0;
  while (x > 1) {
    if (x & 1) return -1;
    res++;
    x = x / 2;
  }
  return res;
}
int main() {
  int i, j;
  long long ans = 0;
  memset(com, -1, sizeof(com));
  cin >> n >> t;
  k = getk(t);
  if (k == -1) {
    printf("0\n");
    return 0;
  }
  k++;
  long long head = 0;
  int rest = k;
  for (i = 60; i >= 0; i--) {
    if (head + (1LL << i) <= n + 1) {
      head += (1LL << i);
      if (rest > 1)
        ans = ans + gc(i, rest);
      else
        ans = ans + gc(i + 1, rest);
      rest--;
    }
    if (rest == 0) break;
  }
  if (t == 1) ans--;
  cout << ans << endl;
  return 0;
}

```