#include <bits/stdc++.h>
using namespace std;
long long n, t;
map<long long, map<int, long long> > mp;
long long gao(long long x, int ns) {
  if (mp[x].find(ns) != mp[x].end()) return mp[x][ns];
  if (ns == 0) return 1;
  if (x == 0) return 0;
  long long lx = x, tx = x;
  while (tx) {
    lx = tx & (-tx);
    tx -= lx;
  }
  return mp[x][ns] = gao(lx - 1, ns) + gao(x - lx, ns - 1);
}
int main() {
  while (cin >> n >> t) {
    if (t == (t & (-t))) {
      int s = 0;
      long long tt = t;
      while (tt) {
        s++;
        tt >>= 1;
      }
      long long ans = gao(n + 1, s);
      if (t == 1) ans--;
      cout << ans << endl;
    } else
      cout << 0 << endl;
  }
  return 0;
}

### Prompt

Please formulate 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 n, t;
map<long long, map<int, long long> > mp;
long long gao(long long x, int ns) {
  if (mp[x].find(ns) != mp[x].end()) return mp[x][ns];
  if (ns == 0) return 1;
  if (x == 0) return 0;
  long long lx = x, tx = x;
  while (tx) {
    lx = tx & (-tx);
    tx -= lx;
  }
  return mp[x][ns] = gao(lx - 1, ns) + gao(x - lx, ns - 1);
}
int main() {
  while (cin >> n >> t) {
    if (t == (t & (-t))) {
      int s = 0;
      long long tt = t;
      while (tt) {
        s++;
        tt >>= 1;
      }
      long long ans = gao(n + 1, s);
      if (t == 1) ans--;
      cout << ans << endl;
    } else
      cout << 0 << endl;
  }
  return 0;
}

```