#include <bits/stdc++.h>
using namespace std;
long long dp[50][2][50];
int main() {
  long long n, t;
  cin >> n >> t;
  n++;
  if (t & (t - 1)) {
    cout << 0 << endl;
    return 0;
  }
  int bit = (int)(log(t) / log(2) + 0.4) + 1;
  vector<int> a;
  for (; n > 0; n >>= 1) a.push_back(n & 1);
  reverse(a.begin(), a.end());
  dp[0][0][0] = 1;
  int m = a.size();
  for (int i = 0; i < m; i++)
    for (int j = 0; j < 2; j++)
      for (int l = 0; l <= i; l++)
        if (dp[i][j][l]) {
          for (int k = 0; k < 2; k++)
            if (!j && k > a[i])
              continue;
            else
              dp[i + 1][j || k < a[i]][l + k] += dp[i][j][l];
        }
  cout << dp[m][0][bit] + dp[m][1][bit] - (bit == 1) << endl;
  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;
long long dp[50][2][50];
int main() {
  long long n, t;
  cin >> n >> t;
  n++;
  if (t & (t - 1)) {
    cout << 0 << endl;
    return 0;
  }
  int bit = (int)(log(t) / log(2) + 0.4) + 1;
  vector<int> a;
  for (; n > 0; n >>= 1) a.push_back(n & 1);
  reverse(a.begin(), a.end());
  dp[0][0][0] = 1;
  int m = a.size();
  for (int i = 0; i < m; i++)
    for (int j = 0; j < 2; j++)
      for (int l = 0; l <= i; l++)
        if (dp[i][j][l]) {
          for (int k = 0; k < 2; k++)
            if (!j && k > a[i])
              continue;
            else
              dp[i + 1][j || k < a[i]][l + k] += dp[i][j][l];
        }
  cout << dp[m][0][bit] + dp[m][1][bit] - (bit == 1) << endl;
  return 0;
}

```