#include <bits/stdc++.h>
using namespace std;
int bit_cnt(long long x) {
  int res = 0;
  for (int i = 0; i < 64; i++) res += (x >> i) & 1;
  return res;
}
bool bit(long long x, int i) { return (x >> i) & 1; }
int main() {
  long long n, t;
  cin >> n >> t;
  if (t & (t - 1)) {
    cout << 0;
    return 0;
  }
  long long C[66][66] = {0};
  C[0][0] = 1;
  for (int i = 1; i < 66; i++) {
    C[i][0] = C[i - 1][0];
    for (int j = 1; j < 66; j++) C[i][j] = C[i - 1][j] + C[i - 1][j - 1];
  }
  int bits = 0;
  for (int i = 0; i < 64; i++)
    if (bit(t, i)) bits = i;
  n++;
  int in_n = 0;
  long long ans = 0;
  for (int i = 63; i >= 0; i--)
    if (bit(n, i)) {
      if (0 <= bits + 1 - in_n && bits + 1 - in_n <= i)
        ans += C[i][bits + 1 - in_n];
      in_n++;
    }
  if (in_n == bits + 1) ans++;
  if (t == 1) ans--;
  cout << ans;
  return 0;
}

### Prompt

Please create a solution in cpp 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;
int bit_cnt(long long x) {
  int res = 0;
  for (int i = 0; i < 64; i++) res += (x >> i) & 1;
  return res;
}
bool bit(long long x, int i) { return (x >> i) & 1; }
int main() {
  long long n, t;
  cin >> n >> t;
  if (t & (t - 1)) {
    cout << 0;
    return 0;
  }
  long long C[66][66] = {0};
  C[0][0] = 1;
  for (int i = 1; i < 66; i++) {
    C[i][0] = C[i - 1][0];
    for (int j = 1; j < 66; j++) C[i][j] = C[i - 1][j] + C[i - 1][j - 1];
  }
  int bits = 0;
  for (int i = 0; i < 64; i++)
    if (bit(t, i)) bits = i;
  n++;
  int in_n = 0;
  long long ans = 0;
  for (int i = 63; i >= 0; i--)
    if (bit(n, i)) {
      if (0 <= bits + 1 - in_n && bits + 1 - in_n <= i)
        ans += C[i][bits + 1 - in_n];
      in_n++;
    }
  if (in_n == bits + 1) ans++;
  if (t == 1) ans--;
  cout << ans;
  return 0;
}

```