#include <bits/stdc++.h>
using namespace std;
int a[44];
long long c[44][44], n, t;
void prepare() {
  c[0][0] = 1;
  for (int i = 0; i <= 40; i++) {
    c[i][0] = 1;
    for (int j = 1; j <= i; j++) c[i][j] = c[i - 1][j] + c[i - 1][j - 1];
  }
}
int main() {
  prepare();
  cin >> n >> t;
  int bit = 0;
  for (int i = 0; i <= 40; i++) {
    if (t == (1ll << i)) {
      bit = i + 1;
      break;
    }
  }
  if (!bit) {
    puts("0");
    return 0;
  }
  int m = 0;
  n++;
  while (n) {
    a[++m] = n % 2, n /= 2;
  }
  long long ans = 0;
  int x = 0;
  for (int i = m; i >= 1; i--) {
    if (a[i]) {
      ans += c[i - 1][bit - x];
    }
    if (a[i]) x++;
    if (x > bit) break;
  }
  if (x == bit) ans++;
  if (bit == 1) ans--;
  cout << ans << endl;
  return 0;
}

### Prompt

Please create a solution in Cpp 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;
int a[44];
long long c[44][44], n, t;
void prepare() {
  c[0][0] = 1;
  for (int i = 0; i <= 40; i++) {
    c[i][0] = 1;
    for (int j = 1; j <= i; j++) c[i][j] = c[i - 1][j] + c[i - 1][j - 1];
  }
}
int main() {
  prepare();
  cin >> n >> t;
  int bit = 0;
  for (int i = 0; i <= 40; i++) {
    if (t == (1ll << i)) {
      bit = i + 1;
      break;
    }
  }
  if (!bit) {
    puts("0");
    return 0;
  }
  int m = 0;
  n++;
  while (n) {
    a[++m] = n % 2, n /= 2;
  }
  long long ans = 0;
  int x = 0;
  for (int i = m; i >= 1; i--) {
    if (a[i]) {
      ans += c[i - 1][bit - x];
    }
    if (a[i]) x++;
    if (x > bit) break;
  }
  if (x == bit) ans++;
  if (bit == 1) ans--;
  cout << ans << endl;
  return 0;
}

```