#include <bits/stdc++.h>
using namespace std;
int main() {
  ios::sync_with_stdio(false), cin.tie(nullptr);
  int64_t n, t;
  cin >> n >> t;
  if (__builtin_popcountll(t) > 1) {
    cout << "0\n";
    return 0;
  }
  int len = 1;
  for (int64_t y = n; y > 0; y >>= 1) len++;
  vector<vector<int64_t>> p(len + 1, vector<int64_t>(len + 1));
  for (int j = 1; j <= len; j++) p[1][j] = 1;
  for (int i = 2; i <= len; i++)
    for (int j = i; j <= len; j++) p[i][j] = p[i][j - 1] + p[i - 1][j - 1];
  n++;
  int x = __builtin_ctzll(t) + 1, counted = 0;
  int64_t ans = __builtin_popcountll(n) == x;
  for (int i = 40; i >= 0; i--) {
    if ((n >> i) & 1) {
      ans += p[x - counted + 1][i + 1];
      counted++;
      if (counted > x) break;
    }
  }
  cout << ans - (x == 1) << '\n';
  return 0;
}

### Prompt

Please formulate 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;
int main() {
  ios::sync_with_stdio(false), cin.tie(nullptr);
  int64_t n, t;
  cin >> n >> t;
  if (__builtin_popcountll(t) > 1) {
    cout << "0\n";
    return 0;
  }
  int len = 1;
  for (int64_t y = n; y > 0; y >>= 1) len++;
  vector<vector<int64_t>> p(len + 1, vector<int64_t>(len + 1));
  for (int j = 1; j <= len; j++) p[1][j] = 1;
  for (int i = 2; i <= len; i++)
    for (int j = i; j <= len; j++) p[i][j] = p[i][j - 1] + p[i - 1][j - 1];
  n++;
  int x = __builtin_ctzll(t) + 1, counted = 0;
  int64_t ans = __builtin_popcountll(n) == x;
  for (int i = 40; i >= 0; i--) {
    if ((n >> i) & 1) {
      ans += p[x - counted + 1][i + 1];
      counted++;
      if (counted > x) break;
    }
  }
  cout << ans - (x == 1) << '\n';
  return 0;
}

```