#include <bits/stdc++.h>
using namespace std;
long long memo[45][45];
long long ncr(int n, int r) {
  if (n < r) return 0;
  if (memo[n][r] != -1) return memo[n][r];
  if (r == 0 || r == n) return memo[n][r] = 1;
  if (r == 1 || r == n - 1) return memo[n][r] = n;
  return memo[n][r] = ncr(n - 1, r - 1) + ncr(n - 1, r);
}
long long solve(long long n, int k) {
  long long x = n;
  int m = 0;
  for (; x; x >>= 1, m++)
    ;
  if ((1ll << m) - 1 == n) return ncr(m, k);
  if (m < k) return 0;
  if (k == 1) return m;
  --m;
  return ncr(m, k) + solve(n & ((1ll << m) - 1), k - 1);
}
int lg(long long t) {
  if (t == 0) return 0;
  return 1 + lg(t >> 1);
}
int main(void) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  memset(memo, -1, sizeof(memo));
  long long n, t;
  cin >> n >> t;
  if (t & (t - 1)) {
    cout << 0 << '\n';
  } else {
    int k = lg(t);
    cout << solve(n + 1, k) - solve(1, k) << '\n';
  }
  return 0;
}

### Prompt

Generate 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 memo[45][45];
long long ncr(int n, int r) {
  if (n < r) return 0;
  if (memo[n][r] != -1) return memo[n][r];
  if (r == 0 || r == n) return memo[n][r] = 1;
  if (r == 1 || r == n - 1) return memo[n][r] = n;
  return memo[n][r] = ncr(n - 1, r - 1) + ncr(n - 1, r);
}
long long solve(long long n, int k) {
  long long x = n;
  int m = 0;
  for (; x; x >>= 1, m++)
    ;
  if ((1ll << m) - 1 == n) return ncr(m, k);
  if (m < k) return 0;
  if (k == 1) return m;
  --m;
  return ncr(m, k) + solve(n & ((1ll << m) - 1), k - 1);
}
int lg(long long t) {
  if (t == 0) return 0;
  return 1 + lg(t >> 1);
}
int main(void) {
  ios::sync_with_stdio(false);
  cin.tie(0);
  memset(memo, -1, sizeof(memo));
  long long n, t;
  cin >> n >> t;
  if (t & (t - 1)) {
    cout << 0 << '\n';
  } else {
    int k = lg(t);
    cout << solve(n + 1, k) - solve(1, k) << '\n';
  }
  return 0;
}

```