#include <bits/stdc++.h>
using namespace std;
map<long long, long long> A[110];
long long n, t, p;
long long solve(long long n, int p) {
  if (p == -1 || n == 0) return 0;
  if (n == 1) return (p == 0);
  if (A[p].find(n) != A[p].end()) return A[p][n];
  long long put = 1;
  while (put * 2 <= n) put <<= 1;
  if (n == put)
    A[p][n] = (p == 0) + solve(n - 1, p);
  else
    A[p][n] = solve(n - put, p - 1) + solve(put, p);
  return A[p][n];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> t;
  n++;
  while (t % 2 == 0) {
    t /= 2;
    ++p;
  }
  if (t != 1) {
    cout << "0";
    return 0;
  }
  cout << solve(n, p) - (p == 0);
  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<long long, long long> A[110];
long long n, t, p;
long long solve(long long n, int p) {
  if (p == -1 || n == 0) return 0;
  if (n == 1) return (p == 0);
  if (A[p].find(n) != A[p].end()) return A[p][n];
  long long put = 1;
  while (put * 2 <= n) put <<= 1;
  if (n == put)
    A[p][n] = (p == 0) + solve(n - 1, p);
  else
    A[p][n] = solve(n - put, p - 1) + solve(put, p);
  return A[p][n];
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cin >> n >> t;
  n++;
  while (t % 2 == 0) {
    t /= 2;
    ++p;
  }
  if (t != 1) {
    cout << "0";
    return 0;
  }
  cout << solve(n, p) - (p == 0);
  return 0;
}

```