#include <bits/stdc++.h>
using namespace std;
int n, m;
int i, j, k, l;
int a[111][111];
map<long long, map<long long, long long> > pre;
long long f(long long n, long long t) {
  if ((t & (t - 1)) || n <= 0) return 0;
  long long p = 1, i = 1;
  while ((i == 1 ? 2 : i * 2 + 2) <= n) {
    p *= 2;
    i = (i == 1 ? 2 : i * 2 + 2);
  }
  long long ans = pre[p][t];
  if (i < n) {
    if (t == 1) ++ans;
    ++i;
  }
  if (i < n) {
    if (t == 2) ++ans;
    ++i;
  }
  ans += f(n - i, t / 2);
  return ans;
}
void dopre() {
  long long n = 2e12;
  long long p, i;
  for (p = 1, i = 1; i <= n; i = i * 2 + 2, p = p * 2) {
    if (p == 1)
      pre[p][1] = 1;
    else if (p == 2)
      pre[p][1] = pre[p][2] = 1;
    else {
      for (typeof((pre[p >> 1]).begin()) it = (pre[p >> 1]).begin();
           it != (pre[p >> 1]).end(); it++) {
        pre[p][it->first] += it->second;
        pre[p][it->first * 2] += it->second;
      }
      pre[p][1]++;
      pre[p][2]++;
    }
  }
}
int main() {
  dopre();
  long long N, T;
  cin >> N >> T;
  cout << f(N, T) << endl;
  return 0;
}

### Prompt

Please create a solution in CPP 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 n, m;
int i, j, k, l;
int a[111][111];
map<long long, map<long long, long long> > pre;
long long f(long long n, long long t) {
  if ((t & (t - 1)) || n <= 0) return 0;
  long long p = 1, i = 1;
  while ((i == 1 ? 2 : i * 2 + 2) <= n) {
    p *= 2;
    i = (i == 1 ? 2 : i * 2 + 2);
  }
  long long ans = pre[p][t];
  if (i < n) {
    if (t == 1) ++ans;
    ++i;
  }
  if (i < n) {
    if (t == 2) ++ans;
    ++i;
  }
  ans += f(n - i, t / 2);
  return ans;
}
void dopre() {
  long long n = 2e12;
  long long p, i;
  for (p = 1, i = 1; i <= n; i = i * 2 + 2, p = p * 2) {
    if (p == 1)
      pre[p][1] = 1;
    else if (p == 2)
      pre[p][1] = pre[p][2] = 1;
    else {
      for (typeof((pre[p >> 1]).begin()) it = (pre[p >> 1]).begin();
           it != (pre[p >> 1]).end(); it++) {
        pre[p][it->first] += it->second;
        pre[p][it->first * 2] += it->second;
      }
      pre[p][1]++;
      pre[p][2]++;
    }
  }
}
int main() {
  dopre();
  long long N, T;
  cin >> N >> T;
  cout << f(N, T) << endl;
  return 0;
}

```