#include <bits/stdc++.h>
using namespace std;
long long ans, C[120][120], n, t, m, i, o, s, j;
int main() {
  scanf("%I64d%I64d", &n, &t);
  for (s = 1, m = 1; s < t; s <<= 1, m++)
    ;
  if (s != t) return puts("0"), 0;
  if (m == 1) ans--;
  n++;
  for (i = 0; i <= 60; ++i)
    for (j = 0; j <= i; ++j) C[i][j] = !j ? 1 : C[i - 1][j - 1] + C[i - 1][j];
  for (i = 60; i >= 0; --i) {
    if (m < 0) break;
    if (o + (1ll << i) <= n) {
      ans += C[i][m];
      m--;
      o += 1ll << i;
    }
  }
  if (!m) ans++;
  printf("%I64d\n", ans);
}

### Prompt

Your challenge is to write 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;
long long ans, C[120][120], n, t, m, i, o, s, j;
int main() {
  scanf("%I64d%I64d", &n, &t);
  for (s = 1, m = 1; s < t; s <<= 1, m++)
    ;
  if (s != t) return puts("0"), 0;
  if (m == 1) ans--;
  n++;
  for (i = 0; i <= 60; ++i)
    for (j = 0; j <= i; ++j) C[i][j] = !j ? 1 : C[i - 1][j - 1] + C[i - 1][j];
  for (i = 60; i >= 0; --i) {
    if (m < 0) break;
    if (o + (1ll << i) <= n) {
      ans += C[i][m];
      m--;
      o += 1ll << i;
    }
  }
  if (!m) ans++;
  printf("%I64d\n", ans);
}

```