#include <bits/stdc++.h>
using namespace std;
int64_t dp[64][64];
int arr[64];
int main() {
  dp[1][1] = dp[1][0] = 1;
  for (int i = 2; i <= 53; i++) {
    for (int j = 1; j <= i; j++) dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j];
    dp[i][0] = dp[i - 1][0];
  }
  int64_t n, t;
  cin >> n >> t;
  n++;
  int co = 0;
  while ((t & 1) == 0) {
    t /= 2;
    co++;
  }
  int64_t ans = 0;
  if (t != 1) {
    cout << ans;
    return 0;
  }
  t = co + 1;
  int num = 0;
  while (n) {
    arr[num++] = n % 2;
    n /= 2;
  }
  int64_t cnt = 0;
  int i;
  for (i = num - 1; i >= 0; i--) {
    if (arr[i]) {
      if (t >= cnt) ans += dp[i][t - cnt];
      cnt++;
    }
  }
  if (t == 1) ans--;
  if (cnt == t) ans++;
  cout << ans;
  return 0;
}

### Prompt

In cpp, your task is to solve 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;
int64_t dp[64][64];
int arr[64];
int main() {
  dp[1][1] = dp[1][0] = 1;
  for (int i = 2; i <= 53; i++) {
    for (int j = 1; j <= i; j++) dp[i][j] = dp[i - 1][j - 1] + dp[i - 1][j];
    dp[i][0] = dp[i - 1][0];
  }
  int64_t n, t;
  cin >> n >> t;
  n++;
  int co = 0;
  while ((t & 1) == 0) {
    t /= 2;
    co++;
  }
  int64_t ans = 0;
  if (t != 1) {
    cout << ans;
    return 0;
  }
  t = co + 1;
  int num = 0;
  while (n) {
    arr[num++] = n % 2;
    n /= 2;
  }
  int64_t cnt = 0;
  int i;
  for (i = num - 1; i >= 0; i--) {
    if (arr[i]) {
      if (t >= cnt) ans += dp[i][t - cnt];
      cnt++;
    }
  }
  if (t == 1) ans--;
  if (cnt == t) ans++;
  cout << ans;
  return 0;
}

```