#include <bits/stdc++.h>
using namespace std;
map<pair<long long, int>, long long> M;
long long getres(long long N, int P) {
  if (N == 0 || P == -1) return 0;
  if (N == 1) return (P == 0);
  if (M.find(make_pair(N, P)) != M.end()) return M[make_pair(N, P)];
  long long pow2 = 1;
  while (pow2 * 2 <= N) pow2 *= 2;
  if (N == pow2)
    M[make_pair(N, P)] = (P == 0) + getres(N - 1, P);
  else
    M[make_pair(N, P)] = getres(pow2, P) + getres(N - pow2, P - 1);
  return M[make_pair(N, P)];
}
int main() {
  long long N, T;
  int P = 0;
  cin >> N >> T;
  ++N;
  while (T % 2 == 0) {
    T /= 2;
    ++P;
  }
  if (T != 1) {
    cout << 0 << '\n';
    return 0;
  }
  cout << getres(N, P) - (P == 0) << '\n';
}

### Prompt

Create a solution in CPP for 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> M;
long long getres(long long N, int P) {
  if (N == 0 || P == -1) return 0;
  if (N == 1) return (P == 0);
  if (M.find(make_pair(N, P)) != M.end()) return M[make_pair(N, P)];
  long long pow2 = 1;
  while (pow2 * 2 <= N) pow2 *= 2;
  if (N == pow2)
    M[make_pair(N, P)] = (P == 0) + getres(N - 1, P);
  else
    M[make_pair(N, P)] = getres(pow2, P) + getres(N - pow2, P - 1);
  return M[make_pair(N, P)];
}
int main() {
  long long N, T;
  int P = 0;
  cin >> N >> T;
  ++N;
  while (T % 2 == 0) {
    T /= 2;
    ++P;
  }
  if (T != 1) {
    cout << 0 << '\n';
    return 0;
  }
  cout << getres(N, P) - (P == 0) << '\n';
}

```