#include <bits/stdc++.h>
using namespace std;
long long n, m, w, sol;
long long tp[63][63];
long long p2[63];
long long solve(long long lg, int nr, int k) {
  if (k < 0 || nr < 0) return 0;
  if (lg >= p2[nr]) return tp[nr][k];
  if (lg <= p2[nr - 1]) return solve(lg, nr - 1, k);
  return solve(lg, nr - 1, k) + solve(lg - p2[nr - 1], nr - 1, k - 1);
}
int main() {
  cin >> n >> m;
  ++n;
  p2[0] = 1;
  for (int i = 1; i < 63; ++i) p2[i] = p2[i - 1] * 2;
  long long x = 1, pt = 0;
  while (x < m) {
    ++pt;
    x *= 2;
  }
  if (x != m) {
    cout << "0\n";
    return 0;
  }
  w = pt;
  tp[0][0] = 1;
  for (int i = 1; i < 63 - 1; ++i) {
    tp[i][0] = 1;
    for (int j = 1; j < 63 - 1; ++j) tp[i][j] = tp[i - 1][j] + tp[i - 1][j - 1];
  }
  x = 1;
  pt = 0;
  while (n > 0) {
    sol += solve(n, pt, w);
    n -= p2[pt];
    ++pt;
  }
  if (m == 1) --sol;
  cout << sol << "\n";
  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;
long long n, m, w, sol;
long long tp[63][63];
long long p2[63];
long long solve(long long lg, int nr, int k) {
  if (k < 0 || nr < 0) return 0;
  if (lg >= p2[nr]) return tp[nr][k];
  if (lg <= p2[nr - 1]) return solve(lg, nr - 1, k);
  return solve(lg, nr - 1, k) + solve(lg - p2[nr - 1], nr - 1, k - 1);
}
int main() {
  cin >> n >> m;
  ++n;
  p2[0] = 1;
  for (int i = 1; i < 63; ++i) p2[i] = p2[i - 1] * 2;
  long long x = 1, pt = 0;
  while (x < m) {
    ++pt;
    x *= 2;
  }
  if (x != m) {
    cout << "0\n";
    return 0;
  }
  w = pt;
  tp[0][0] = 1;
  for (int i = 1; i < 63 - 1; ++i) {
    tp[i][0] = 1;
    for (int j = 1; j < 63 - 1; ++j) tp[i][j] = tp[i - 1][j] + tp[i - 1][j - 1];
  }
  x = 1;
  pt = 0;
  while (n > 0) {
    sol += solve(n, pt, w);
    n -= p2[pt];
    ++pt;
  }
  if (m == 1) --sol;
  cout << sol << "\n";
  return 0;
}

```