#include <bits/stdc++.h>
using namespace std;
long long c[55][55], p[55], ans, n, x, t, m = 45;
int main() {
  for (long long i = 0; i < m; i++)
    for (long long j = 0; j <= i; j++)
      c[i][j] = i && j ? c[i - 1][j] + c[i - 1][j - 1] : 1;
  for (long long i = 0; i < m; i++) p[i] = 1LL << i;
  cin >> n >> t, n++;
  if ((t & -t) != t) return puts("0"), 0;
  for (long long i = 0; i < m; i++)
    if (t & 1LL << i) {
      x = i;
      break;
    }
  for (long long i = 0; i < m && p[i] <= n; i++) ans += c[i][x], n -= p[i];
  for (long long i = m - 1; i >= 0 && x >= 0; i--)
    if (p[i] <= n) ans += c[i][x], n -= p[i], x--;
  cout << ans - (t == 1) << endl;
  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;
long long c[55][55], p[55], ans, n, x, t, m = 45;
int main() {
  for (long long i = 0; i < m; i++)
    for (long long j = 0; j <= i; j++)
      c[i][j] = i && j ? c[i - 1][j] + c[i - 1][j - 1] : 1;
  for (long long i = 0; i < m; i++) p[i] = 1LL << i;
  cin >> n >> t, n++;
  if ((t & -t) != t) return puts("0"), 0;
  for (long long i = 0; i < m; i++)
    if (t & 1LL << i) {
      x = i;
      break;
    }
  for (long long i = 0; i < m && p[i] <= n; i++) ans += c[i][x], n -= p[i];
  for (long long i = m - 1; i >= 0 && x >= 0; i--)
    if (p[i] <= n) ans += c[i][x], n -= p[i], x--;
  cout << ans - (t == 1) << endl;
  return 0;
}

```