#include <bits/stdc++.h>
using namespace std;
long long s[50][50];
long long foo(long long n, int i, int j) {
  long long q = (1LL << i), p = (1LL << j);
  if (i < j || j < 0) return 0;
  if (n == q) return s[i][j];
  if (n <= q / 2) return foo(n, i - 1, j);
  return foo(q / 2, i - 1, j) + foo(n - q / 2, i - 1, j - 1);
}
int main() {
  long long n, t;
  cin >> n >> t;
  n++;
  if (t & (t - 1)) {
    cout << 0 << endl;
    return 0;
  }
  for (int i = 0; i < 50; i++) s[i][i] = s[i][0] = 1;
  for (int i = 1; i < 50; i++)
    for (int j = 1; j < i; j++) s[i][j] = s[i - 1][j] + s[i - 1][j - 1];
  long long res = 0;
  int i, j = 0;
  while (t >>= 1) j++;
  for (i = 0;; i++) {
    long long q = (1LL << i);
    if (n - q < 0) break;
    res += foo(q, i, j);
    n -= q;
  }
  res += foo(n, i, j);
  if (j == 0) res--;
  cout << res;
  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;
long long s[50][50];
long long foo(long long n, int i, int j) {
  long long q = (1LL << i), p = (1LL << j);
  if (i < j || j < 0) return 0;
  if (n == q) return s[i][j];
  if (n <= q / 2) return foo(n, i - 1, j);
  return foo(q / 2, i - 1, j) + foo(n - q / 2, i - 1, j - 1);
}
int main() {
  long long n, t;
  cin >> n >> t;
  n++;
  if (t & (t - 1)) {
    cout << 0 << endl;
    return 0;
  }
  for (int i = 0; i < 50; i++) s[i][i] = s[i][0] = 1;
  for (int i = 1; i < 50; i++)
    for (int j = 1; j < i; j++) s[i][j] = s[i - 1][j] + s[i - 1][j - 1];
  long long res = 0;
  int i, j = 0;
  while (t >>= 1) j++;
  for (i = 0;; i++) {
    long long q = (1LL << i);
    if (n - q < 0) break;
    res += foo(q, i, j);
    n -= q;
  }
  res += foo(n, i, j);
  if (j == 0) res--;
  cout << res;
  return 0;
}

```