#include <bits/stdc++.h>
using namespace std;
const long long INF = 1000000000000000000;
map<pair<long long, int>, long long> memo;
long long solve(long long n, int x) {
  pair<long long, int> p = make_pair(n, x);
  if (memo.find(p) != memo.end()) return memo[p];
  if (n == 0) return memo[p] = 0;
  if (n == 1) return memo[p] = (x == 0) ? 1 : 0;
  return memo[p] = solve((n + 1) / 2, x) + (x ? solve(n / 2, x - 1) : 0);
}
int main() {
  ios_base::sync_with_stdio(false);
  long long n, t;
  cin >> n >> t;
  for (int i = 0; (1ll << i) <= t; ++i)
    if (t == (1ll << i)) {
      cout << solve(n + 2, i + 1) - solve(2, i + 1);
      return 0;
    }
  cout << 0;
  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;
const long long INF = 1000000000000000000;
map<pair<long long, int>, long long> memo;
long long solve(long long n, int x) {
  pair<long long, int> p = make_pair(n, x);
  if (memo.find(p) != memo.end()) return memo[p];
  if (n == 0) return memo[p] = 0;
  if (n == 1) return memo[p] = (x == 0) ? 1 : 0;
  return memo[p] = solve((n + 1) / 2, x) + (x ? solve(n / 2, x - 1) : 0);
}
int main() {
  ios_base::sync_with_stdio(false);
  long long n, t;
  cin >> n >> t;
  for (int i = 0; (1ll << i) <= t; ++i)
    if (t == (1ll << i)) {
      cout << solve(n + 2, i + 1) - solve(2, i + 1);
      return 0;
    }
  cout << 0;
  return 0;
}

```