#include <bits/stdc++.h>
using namespace std;
int digi[61];
long long F[61][61][2];
int wantOne;
long long f(int cur, int one, int alreadyLess) {
  if (cur == 60) return (one == wantOne);
  long long &ret = F[cur][one][alreadyLess];
  if (ret != -1) return ret;
  ret = 0;
  for (int i = 0; i <= 1; i++)
    if (alreadyLess || (i <= digi[cur]))
      ret += f(cur + 1, one + i, alreadyLess | (i < digi[cur]));
  return ret;
}
int MAIN() {
  long long n, t;
  while (cin >> n >> t) {
    int cnt = 1;
    while (t % 2 == 0) {
      cnt++;
      t /= 2;
    }
    if (t != 1)
      cout << 0 << endl;
    else {
      n++;
      long long ans = 0;
      if (cnt == 1) ans--;
      wantOne = cnt;
      memset(F, 0xff, sizeof(F));
      for (int i = 0; i <= 59; i++) {
        digi[59 - i] = n % 2;
        n /= 2;
      }
      cout << ans + f(0, 0, 0) << endl;
    }
  }
  return 0;
}
int main() {
  ios ::sync_with_stdio(false);
  cout << fixed << setprecision(16);
  return MAIN();
}

### Prompt

Please formulate 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;
int digi[61];
long long F[61][61][2];
int wantOne;
long long f(int cur, int one, int alreadyLess) {
  if (cur == 60) return (one == wantOne);
  long long &ret = F[cur][one][alreadyLess];
  if (ret != -1) return ret;
  ret = 0;
  for (int i = 0; i <= 1; i++)
    if (alreadyLess || (i <= digi[cur]))
      ret += f(cur + 1, one + i, alreadyLess | (i < digi[cur]));
  return ret;
}
int MAIN() {
  long long n, t;
  while (cin >> n >> t) {
    int cnt = 1;
    while (t % 2 == 0) {
      cnt++;
      t /= 2;
    }
    if (t != 1)
      cout << 0 << endl;
    else {
      n++;
      long long ans = 0;
      if (cnt == 1) ans--;
      wantOne = cnt;
      memset(F, 0xff, sizeof(F));
      for (int i = 0; i <= 59; i++) {
        digi[59 - i] = n % 2;
        n /= 2;
      }
      cout << ans + f(0, 0, 0) << endl;
    }
  }
  return 0;
}
int main() {
  ios ::sync_with_stdio(false);
  cout << fixed << setprecision(16);
  return MAIN();
}

```