#include <bits/stdc++.h>
using namespace std;
int Popcount(long long int a) {
  int k;
  for (k = 0; a; a >>= 1)
    if (a & 1ll) k++;
  return k;
}
int Ffs(long long int a) {
  int k;
  for (k = 0; a; a >>= 1, k++)
    ;
  return k;
}
long long int C[50][50];
long long int getAns(long long int n, long long int t) {
  if (n < t || t <= 0) return 0ll;
  long long int i;
  int k = 0, j = Ffs(t) - 1;
  for (i = 1ll; (i << 1) < n; (i <<= 1), k++)
    ;
  long long int res = getAns(n - i, t >> 1);
  if (k < j) return res;
  return C[k][j] + res;
}
int main() {
  C[0][0] = 1ll;
  for (int i = 1; i < 50; i++) {
    C[i][0] = C[i][i] = 1ll;
    for (int j = 1; j < i; j++) C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
  }
  long long int N, T;
  while (cin >> N >> T) {
    N++;
    if (Popcount(T) != 1) {
      puts("0");
      continue;
    }
    long long int Ans = 0ll, x, di = 0ll;
    for (x = 1ll; x + di <= N; x <<= 1) {
      Ans += getAns(x, T);
      di += x;
    }
    Ans += getAns(N - di, T);
    if (T == 1ll) Ans--;
    cout << Ans << endl;
  }
  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;
int Popcount(long long int a) {
  int k;
  for (k = 0; a; a >>= 1)
    if (a & 1ll) k++;
  return k;
}
int Ffs(long long int a) {
  int k;
  for (k = 0; a; a >>= 1, k++)
    ;
  return k;
}
long long int C[50][50];
long long int getAns(long long int n, long long int t) {
  if (n < t || t <= 0) return 0ll;
  long long int i;
  int k = 0, j = Ffs(t) - 1;
  for (i = 1ll; (i << 1) < n; (i <<= 1), k++)
    ;
  long long int res = getAns(n - i, t >> 1);
  if (k < j) return res;
  return C[k][j] + res;
}
int main() {
  C[0][0] = 1ll;
  for (int i = 1; i < 50; i++) {
    C[i][0] = C[i][i] = 1ll;
    for (int j = 1; j < i; j++) C[i][j] = C[i - 1][j - 1] + C[i - 1][j];
  }
  long long int N, T;
  while (cin >> N >> T) {
    N++;
    if (Popcount(T) != 1) {
      puts("0");
      continue;
    }
    long long int Ans = 0ll, x, di = 0ll;
    for (x = 1ll; x + di <= N; x <<= 1) {
      Ans += getAns(x, T);
      di += x;
    }
    Ans += getAns(N - di, T);
    if (T == 1ll) Ans--;
    cout << Ans << endl;
  }
  return 0;
}

```