#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:16777216")
using namespace std;
const double PI = 2.0 * acos(0.0);
template <class T>
inline T MAX(T a, T b) {
  if (a > b) return a;
  return b;
}
template <class T>
inline T MIN(T a, T b) {
  if (a < b) return a;
  return b;
}
int nbit, a[111];
long long n, t, f[111][111][3];
int main() {
  cin >> n >> t;
  n++;
  while (n) {
    a[++nbit] = n % 2;
    n /= 2;
  }
  int need = -1;
  for (int i = (0), _b = (60); i <= _b; i++)
    if ((1LL << i) == t) need = i + 1;
  if (need == -1) {
    cout << 0;
    return 0;
  }
  f[nbit + 1][0][0] = 1;
  for (int i = (nbit + 1), _b = (1); i >= _b; i--)
    for (int nbit1 = (0), _b = (nbit); nbit1 <= _b; nbit1++)
      for (int sm = (0), _b = (1); sm <= _b; sm++) {
        f[i - 1][nbit1][sm | (0 < a[i - 1])] += f[i][nbit1][sm];
        if (sm == 1 || (sm == 0 && a[i - 1] == 1))
          f[i - 1][nbit1 + 1][sm] += f[i][nbit1][sm];
      }
  long long res = f[1][need][0] + f[1][need][1];
  if (need == 1) res--;
  cout << res;
}

### 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>
#pragma comment(linker, "/STACK:16777216")
using namespace std;
const double PI = 2.0 * acos(0.0);
template <class T>
inline T MAX(T a, T b) {
  if (a > b) return a;
  return b;
}
template <class T>
inline T MIN(T a, T b) {
  if (a < b) return a;
  return b;
}
int nbit, a[111];
long long n, t, f[111][111][3];
int main() {
  cin >> n >> t;
  n++;
  while (n) {
    a[++nbit] = n % 2;
    n /= 2;
  }
  int need = -1;
  for (int i = (0), _b = (60); i <= _b; i++)
    if ((1LL << i) == t) need = i + 1;
  if (need == -1) {
    cout << 0;
    return 0;
  }
  f[nbit + 1][0][0] = 1;
  for (int i = (nbit + 1), _b = (1); i >= _b; i--)
    for (int nbit1 = (0), _b = (nbit); nbit1 <= _b; nbit1++)
      for (int sm = (0), _b = (1); sm <= _b; sm++) {
        f[i - 1][nbit1][sm | (0 < a[i - 1])] += f[i][nbit1][sm];
        if (sm == 1 || (sm == 0 && a[i - 1] == 1))
          f[i - 1][nbit1 + 1][sm] += f[i][nbit1][sm];
      }
  long long res = f[1][need][0] + f[1][need][1];
  if (need == 1) res--;
  cout << res;
}

```