#include <bits/stdc++.h>
using namespace std;
long long pascal[77][77];
void fill_pascal() {
  for (int n = 0; n < 77; ++n) {
    pascal[n][0] = 1;
    for (int k = 1; k <= n; ++k)
      pascal[n][k] = pascal[n - 1][k - 1] + pascal[n - 1][k];
  }
}
int main() {
  fill_pascal();
  long long T, N;
  scanf("%I64d%I64d", &N, &T);
  long long l, k;
  for (l = 0; (1ll << l) < T; l++)
    ;
  for (k = 0; (1ll << k) < N; k++)
    ;
  if ((1ll << l) != T) {
    printf("0\n");
    return 0;
  }
  l++;
  k += 2;
  long long V = N + 2;
  long long r = 0;
  int add = 0;
  for (; k >= 0; --k) {
    if (V & (1ll << k)) {
      if (l - add >= 0) r += pascal[k][l - add];
      ++add;
    }
  }
  if (l == 1) --r;
  printf("%I64d\n", r);
  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;
long long pascal[77][77];
void fill_pascal() {
  for (int n = 0; n < 77; ++n) {
    pascal[n][0] = 1;
    for (int k = 1; k <= n; ++k)
      pascal[n][k] = pascal[n - 1][k - 1] + pascal[n - 1][k];
  }
}
int main() {
  fill_pascal();
  long long T, N;
  scanf("%I64d%I64d", &N, &T);
  long long l, k;
  for (l = 0; (1ll << l) < T; l++)
    ;
  for (k = 0; (1ll << k) < N; k++)
    ;
  if ((1ll << l) != T) {
    printf("0\n");
    return 0;
  }
  l++;
  k += 2;
  long long V = N + 2;
  long long r = 0;
  int add = 0;
  for (; k >= 0; --k) {
    if (V & (1ll << k)) {
      if (l - add >= 0) r += pascal[k][l - add];
      ++add;
    }
  }
  if (l == 1) --r;
  printf("%I64d\n", r);
  return 0;
}

```