#include <bits/stdc++.h>
using namespace std;
map<pair<long long, int>, long long> mp;
long long func(long long n, int x) {
  pair<long long, int> p = make_pair(n, x);
  if (mp.find(p) != mp.end()) return mp[p];
  if (n == 0) return mp[p] = 0;
  if (n == 1) return mp[p] = ((x == 0) ? 1 : 0);
  long long ans = 0;
  ans += func((n + 1) / 2, x);
  if (x != 0) ans += func(n / 2, x - 1);
  return mp[p] = ans;
}
int main(void) {
  long long n, t;
  int i;
  cin >> n >> t;
  for ((i) = 0; (i) < (int)(50); (i)++)
    if (t == (1ll << i)) {
      long long ans = func(n + 2, i + 1) - func(2, i + 1);
      cout << ans << endl;
      return 0;
    }
  cout << 0 << endl;
  return 0;
}

### Prompt

Your challenge is to write 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;
map<pair<long long, int>, long long> mp;
long long func(long long n, int x) {
  pair<long long, int> p = make_pair(n, x);
  if (mp.find(p) != mp.end()) return mp[p];
  if (n == 0) return mp[p] = 0;
  if (n == 1) return mp[p] = ((x == 0) ? 1 : 0);
  long long ans = 0;
  ans += func((n + 1) / 2, x);
  if (x != 0) ans += func(n / 2, x - 1);
  return mp[p] = ans;
}
int main(void) {
  long long n, t;
  int i;
  cin >> n >> t;
  for ((i) = 0; (i) < (int)(50); (i)++)
    if (t == (1ll << i)) {
      long long ans = func(n + 2, i + 1) - func(2, i + 1);
      cout << ans << endl;
      return 0;
    }
  cout << 0 << endl;
  return 0;
}

```