#include <bits/stdc++.h>
using namespace std;
template <class T>
T abs(T x) {
  return x > 0 ? x : -x;
}
long long n, m;
map<pair<long long, int>, long long> all;
long long calc(long long len, long long cur, int k) {
  if (cur <= 0 || k < 0) return 0;
  if (len == 1) return int(k == 0);
  if (len == cur && all.find(make_pair(len, k)) != all.end())
    return all[make_pair(len, k)];
  long long ans =
      calc(len / 2, min(cur, len / 2), k) + calc(len / 2, cur - len / 2, k - 1);
  if (len == cur) all[make_pair(len, k)] = ans;
  return ans;
}
int main() {
  cin >> n >> m;
  int k = 1;
  while (m % 2 == 0) {
    k++;
    m /= 2;
  }
  if (m != 1) {
    printf("0\n");
    return 0;
  }
  n += 2;
  long long cur = 1;
  while (cur <= n) cur *= 2;
  printf("%I64d\n", calc(cur, n, k) - int(k == 1));
  return 0;
}

### Prompt

Your challenge is to write 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;
template <class T>
T abs(T x) {
  return x > 0 ? x : -x;
}
long long n, m;
map<pair<long long, int>, long long> all;
long long calc(long long len, long long cur, int k) {
  if (cur <= 0 || k < 0) return 0;
  if (len == 1) return int(k == 0);
  if (len == cur && all.find(make_pair(len, k)) != all.end())
    return all[make_pair(len, k)];
  long long ans =
      calc(len / 2, min(cur, len / 2), k) + calc(len / 2, cur - len / 2, k - 1);
  if (len == cur) all[make_pair(len, k)] = ans;
  return ans;
}
int main() {
  cin >> n >> m;
  int k = 1;
  while (m % 2 == 0) {
    k++;
    m /= 2;
  }
  if (m != 1) {
    printf("0\n");
    return 0;
  }
  n += 2;
  long long cur = 1;
  while (cur <= n) cur *= 2;
  printf("%I64d\n", calc(cur, n, k) - int(k == 1));
  return 0;
}

```