#include <bits/stdc++.h>
using namespace std;
long long DP[51][51];
int Log(long long N) {
  long long S = 1;
  int Ans = 0;
  while (S < N) {
    S *= 2;
    Ans++;
  }
  return (N == S) ? Ans : -1;
}
long long Solve(int Level, long long Len, int T) {
  if (!Len || T < 0) return 0;
  long long Total = 1LL << Level;
  if (Len < Total / 2) return Solve(Level - 1, Len, T);
  return DP[Level - 1][T] + Solve(Level - 1, Len - Total / 2, T - 1);
}
int main() {
  long long N, M;
  cin >> N >> M;
  int T = Log(M);
  if (T == -1) {
    printf("0\n");
    return 0;
  }
  DP[0][0] = 1;
  for (int i = 0; i < 50; i++)
    for (int j = 0; j <= 50; j++) {
      DP[i + 1][j] += DP[i][j];
      if (j < 50) DP[i + 1][j + 1] += DP[i][j];
    }
  long long Len = 2, Ans = 0;
  int Level = 1;
  while (N >= Len) {
    Ans += DP[Level][T];
    N -= Len;
    Len *= 2;
    Level++;
  }
  Ans += Solve(Level, N, T);
  cout << Ans << endl;
  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 DP[51][51];
int Log(long long N) {
  long long S = 1;
  int Ans = 0;
  while (S < N) {
    S *= 2;
    Ans++;
  }
  return (N == S) ? Ans : -1;
}
long long Solve(int Level, long long Len, int T) {
  if (!Len || T < 0) return 0;
  long long Total = 1LL << Level;
  if (Len < Total / 2) return Solve(Level - 1, Len, T);
  return DP[Level - 1][T] + Solve(Level - 1, Len - Total / 2, T - 1);
}
int main() {
  long long N, M;
  cin >> N >> M;
  int T = Log(M);
  if (T == -1) {
    printf("0\n");
    return 0;
  }
  DP[0][0] = 1;
  for (int i = 0; i < 50; i++)
    for (int j = 0; j <= 50; j++) {
      DP[i + 1][j] += DP[i][j];
      if (j < 50) DP[i + 1][j + 1] += DP[i][j];
    }
  long long Len = 2, Ans = 0;
  int Level = 1;
  while (N >= Len) {
    Ans += DP[Level][T];
    N -= Len;
    Len *= 2;
    Level++;
  }
  Ans += Solve(Level, N, T);
  cout << Ans << endl;
  return 0;
}

```