#include <bits/stdc++.h>
using namespace std;
long long n, t;
long long c[60][60];
long long solve(int logn, long long n, int k, int can) {
  if (k < 0) return 0;
  if (logn == 1) return can && (k == 0 || (k == 1 && n == 2));
  if ((1LL << (logn - 1)) >= n) return solve(logn - 1, n, k, 1);
  return c[logn - 1][k] + solve(logn - 1, n - (1LL << (logn - 1)), k - 1, 0);
}
int main() {
  c[1][0] = c[1][1] = 1;
  for (int i = 2; i < 60; i++) {
    for (int j = 1; j <= i; j++) c[i][j] += c[i - 1][j - 1];
    c[i][2]--;
    c[i][0]++;
    for (int j = 0; j <= i; j++) c[i][j] += c[i - 1][j];
  }
  cin >> n >> t;
  if (t & (t - 1)) {
    puts("0");
    return 0;
  }
  int k = 0;
  for (; (1LL << k) < t; k++)
    ;
  cout << solve(50, n, k, 1) << endl;
  return 0;
}

### Prompt

Your task is to create 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 n, t;
long long c[60][60];
long long solve(int logn, long long n, int k, int can) {
  if (k < 0) return 0;
  if (logn == 1) return can && (k == 0 || (k == 1 && n == 2));
  if ((1LL << (logn - 1)) >= n) return solve(logn - 1, n, k, 1);
  return c[logn - 1][k] + solve(logn - 1, n - (1LL << (logn - 1)), k - 1, 0);
}
int main() {
  c[1][0] = c[1][1] = 1;
  for (int i = 2; i < 60; i++) {
    for (int j = 1; j <= i; j++) c[i][j] += c[i - 1][j - 1];
    c[i][2]--;
    c[i][0]++;
    for (int j = 0; j <= i; j++) c[i][j] += c[i - 1][j];
  }
  cin >> n >> t;
  if (t & (t - 1)) {
    puts("0");
    return 0;
  }
  int k = 0;
  for (; (1LL << k) < t; k++)
    ;
  cout << solve(50, n, k, 1) << endl;
  return 0;
}

```