#include <bits/stdc++.h>
using namespace std;
int main() {
  long long n, t;
  cin >> n >> t;
  int d = 1;
  while (t > 1) {
    if (t % 2) {
      cout << 0 << endl;
      return 0;
    }
    t /= 2;
    d++;
  }
  long long dp[60][60] = {};
  dp[1][1] = 1;
  long long cur = 1;
  long long lv = 1;
  long long ans = 0;
  int i = 2;
  n++;
  for (i = 2; i <= 60; i++) {
    for (int j = 1; j <= i; j++) {
      dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j];
    }
    lv *= 2;
    cur += lv;
    if (cur > n) break;
    ans += dp[i][d];
  }
  n -= lv - 1;
  while (lv > 1) {
    lv /= 2;
    i--;
    if (n >= lv) {
      ans += dp[i][d];
      n -= lv;
      d--;
    }
  }
  cout << ans << endl;
  return 0;
}

### 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;
int main() {
  long long n, t;
  cin >> n >> t;
  int d = 1;
  while (t > 1) {
    if (t % 2) {
      cout << 0 << endl;
      return 0;
    }
    t /= 2;
    d++;
  }
  long long dp[60][60] = {};
  dp[1][1] = 1;
  long long cur = 1;
  long long lv = 1;
  long long ans = 0;
  int i = 2;
  n++;
  for (i = 2; i <= 60; i++) {
    for (int j = 1; j <= i; j++) {
      dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j];
    }
    lv *= 2;
    cur += lv;
    if (cur > n) break;
    ans += dp[i][d];
  }
  n -= lv - 1;
  while (lv > 1) {
    lv /= 2;
    i--;
    if (n >= lv) {
      ans += dp[i][d];
      n -= lv;
      d--;
    }
  }
  cout << ans << endl;
  return 0;
}

```