#include <bits/stdc++.h>
using namespace std;
vector<vector<long long> > ans(100 + 47);
long long Ans(long long N, int P) {
  if (N == 1) return (int)(P == 0);
  long long a = 1;
  int p = 0;
  while (a - 1 <= N) {
    a *= 2;
    p++;
  }
  if (a - 1 > N) {
    a /= 2;
    p--;
  }
  long long ret = ans[p][P];
  if (a - 1 == N) return ret;
  if (a <= N && P == 0) ret++;
  if (a == N) return ret;
  if (P > 0) ret += Ans(N - a, P - 1);
  return ret;
}
int main() {
  long long N, T;
  int P = 0;
  scanf(" %I64d %I64d", &N, &T);
  while (T > 1) {
    if (T % 2 != 0) {
      printf("0\n");
      return 0;
    }
    P++;
    T /= 2;
  }
  ans[1].resize(100 + 47, 0);
  ans[1][0] = 1;
  for (int i = 1; i < 100; i++) {
    ans[i + 1] = ans[i];
    for (int j = 0; j < 100; j++) ans[i + 1][j + 1] += ans[i][j];
    ans[i + 1][0]++;
  }
  N++;
  printf("%I64d\n", Ans(N, P) - (int)(P == 0));
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
vector<vector<long long> > ans(100 + 47);
long long Ans(long long N, int P) {
  if (N == 1) return (int)(P == 0);
  long long a = 1;
  int p = 0;
  while (a - 1 <= N) {
    a *= 2;
    p++;
  }
  if (a - 1 > N) {
    a /= 2;
    p--;
  }
  long long ret = ans[p][P];
  if (a - 1 == N) return ret;
  if (a <= N && P == 0) ret++;
  if (a == N) return ret;
  if (P > 0) ret += Ans(N - a, P - 1);
  return ret;
}
int main() {
  long long N, T;
  int P = 0;
  scanf(" %I64d %I64d", &N, &T);
  while (T > 1) {
    if (T % 2 != 0) {
      printf("0\n");
      return 0;
    }
    P++;
    T /= 2;
  }
  ans[1].resize(100 + 47, 0);
  ans[1][0] = 1;
  for (int i = 1; i < 100; i++) {
    ans[i + 1] = ans[i];
    for (int j = 0; j < 100; j++) ans[i + 1][j + 1] += ans[i][j];
    ans[i + 1][0]++;
  }
  N++;
  printf("%I64d\n", Ans(N, P) - (int)(P == 0));
  return 0;
}

```