#include <bits/stdc++.h>
using namespace std;
enum { MAXK = 50 };
long long m[MAXK][MAXK];
long long solve(long long v, int p) {
  if (v == 0)
    return 0;
  else if (p == 0)
    return 1;
  else {
    int k = log2(v);
    return m[k][p] + solve(v - (1LL << k), p - 1);
  }
}
int main() {
  long long n, t;
  cin >> n >> t;
  if (t & (t - 1))
    puts("0");
  else {
    int p;
    for (p = 0; t >> p; p++)
      ;
    --p;
    long long ans = 0;
    memset(m, 0, sizeof(m));
    m[0][0] = 1;
    for (int i = 1; n >= (1LL << i); i++) {
      for (int j = p; j > 0; j--) m[i][j] = m[i - 1][j] + m[i - 1][j - 1];
      m[i][0] = 1;
      ans += m[i][p];
      n -= (1LL << i);
    }
    cout << ans + solve(n, p) << endl;
  }
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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;
enum { MAXK = 50 };
long long m[MAXK][MAXK];
long long solve(long long v, int p) {
  if (v == 0)
    return 0;
  else if (p == 0)
    return 1;
  else {
    int k = log2(v);
    return m[k][p] + solve(v - (1LL << k), p - 1);
  }
}
int main() {
  long long n, t;
  cin >> n >> t;
  if (t & (t - 1))
    puts("0");
  else {
    int p;
    for (p = 0; t >> p; p++)
      ;
    --p;
    long long ans = 0;
    memset(m, 0, sizeof(m));
    m[0][0] = 1;
    for (int i = 1; n >= (1LL << i); i++) {
      for (int j = p; j > 0; j--) m[i][j] = m[i - 1][j] + m[i - 1][j - 1];
      m[i][0] = 1;
      ans += m[i][p];
      n -= (1LL << i);
    }
    cout << ans + solve(n, p) << endl;
  }
}

```