#include <bits/stdc++.h>
using namespace std;
const int maxn = (int)2e5;
const int inf = (int)1e9;
long long ans;
long long a[100][100];
int k;
void solve(long long t, int i) {
  for (int kof = 0; t > 0; i--) {
    if (1ll << i <= t) {
      t -= (1ll << i);
      if (k - kof < 0) return;
      ans += a[i][k - kof];
      kof++;
    }
  }
}
int main() {
  int i;
  long long n, t, d;
  cin >> n >> t;
  n++;
  for (k = 0; (t & 1) == 0; k++, t >>= 1)
    ;
  if (t != 1) {
    printf("0\n");
    return 0;
  }
  d = 2;
  a[0][0] = 1;
  for (i = 0; d * 2 - 1 <= n; d *= 2, i++) {
    for (int j = 0; j <= i; j++) a[i + 1][j + 1] = a[i][j];
    for (int j = 0; j <= i + 1; j++) a[i + 1][j] += a[i][j];
    ans += a[i + 1][k];
  }
  solve(n - (d - 1), i);
  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;
const int maxn = (int)2e5;
const int inf = (int)1e9;
long long ans;
long long a[100][100];
int k;
void solve(long long t, int i) {
  for (int kof = 0; t > 0; i--) {
    if (1ll << i <= t) {
      t -= (1ll << i);
      if (k - kof < 0) return;
      ans += a[i][k - kof];
      kof++;
    }
  }
}
int main() {
  int i;
  long long n, t, d;
  cin >> n >> t;
  n++;
  for (k = 0; (t & 1) == 0; k++, t >>= 1)
    ;
  if (t != 1) {
    printf("0\n");
    return 0;
  }
  d = 2;
  a[0][0] = 1;
  for (i = 0; d * 2 - 1 <= n; d *= 2, i++) {
    for (int j = 0; j <= i; j++) a[i + 1][j + 1] = a[i][j];
    for (int j = 0; j <= i + 1; j++) a[i + 1][j] += a[i][j];
    ans += a[i + 1][k];
  }
  solve(n - (d - 1), i);
  cout << ans << endl;
  return 0;
}

```