#include <bits/stdc++.h>
using namespace std;
map<pair<long long int, int>, long long int> memo;
map<pair<long long int, int>, long long int> visitado;
long long int f(long long int n, long long int e) {
  if (n == 0 && e == 0) return 1;
  if (n == 0) return 0;
  if (e == 1 && n == 1) return 1;
  if (e == 0) return 1;
  if (visitado[make_pair(n, e)] == 1) return memo[make_pair(n, e)];
  visitado[make_pair(n, e)] = 1;
  return memo[make_pair(n, e)] = f((n - 1) / 2, e - 1) + f(n / 2, e);
}
int main() {
  long long int n, t;
  scanf("%I64d%I64d", &n, &t);
  n++;
  int e = 0;
  while (t > 1) {
    if (t % 2 == 0) {
      t /= 2;
      e++;
    } else {
      e = -1;
      t = 1;
    }
  }
  if (e == -1)
    printf("0\n");
  else {
    long long int resp = f(n, e + 1);
    if (e == 0) resp--;
    printf("%I64d\n", resp);
  }
  return 0;
}

### Prompt

Create a solution in CPP for 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;
map<pair<long long int, int>, long long int> memo;
map<pair<long long int, int>, long long int> visitado;
long long int f(long long int n, long long int e) {
  if (n == 0 && e == 0) return 1;
  if (n == 0) return 0;
  if (e == 1 && n == 1) return 1;
  if (e == 0) return 1;
  if (visitado[make_pair(n, e)] == 1) return memo[make_pair(n, e)];
  visitado[make_pair(n, e)] = 1;
  return memo[make_pair(n, e)] = f((n - 1) / 2, e - 1) + f(n / 2, e);
}
int main() {
  long long int n, t;
  scanf("%I64d%I64d", &n, &t);
  n++;
  int e = 0;
  while (t > 1) {
    if (t % 2 == 0) {
      t /= 2;
      e++;
    } else {
      e = -1;
      t = 1;
    }
  }
  if (e == -1)
    printf("0\n");
  else {
    long long int resp = f(n, e + 1);
    if (e == 0) resp--;
    printf("%I64d\n", resp);
  }
  return 0;
}

```