#include <bits/stdc++.h>
using namespace std;
long long n, t;
long long c[62][62];
int main() {
  cin >> n >> t;
  if (t != 1 && (t & (t - 1))) {
    cout << 0;
    return 0;
  }
  c[0][0] = 1;
  for (int i = 1; i <= 60; i++) {
    c[i][0] = 1;
    for (int j = 1; j <= i; j++) c[i][j] = c[i - 1][j] + c[i - 1][j - 1];
  }
  int p;
  for (p = 0; (1LL << p) < t; p++)
    ;
  p++;
  long long res = 0;
  n += 2;
  for (int i = 60; i >= 0; i--) {
    if ((1LL << (i)) & n) res += c[i][p];
    p -= ((1LL << (i)) & n) > 0;
    if (p < 0) break;
  }
  if (t == 1 && n == 1) res++;
  if (t == 1 && n != 1) res--;
  cout << res << endl;
  return 0;
}

### Prompt

Your task is to create 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 n, t;
long long c[62][62];
int main() {
  cin >> n >> t;
  if (t != 1 && (t & (t - 1))) {
    cout << 0;
    return 0;
  }
  c[0][0] = 1;
  for (int i = 1; i <= 60; i++) {
    c[i][0] = 1;
    for (int j = 1; j <= i; j++) c[i][j] = c[i - 1][j] + c[i - 1][j - 1];
  }
  int p;
  for (p = 0; (1LL << p) < t; p++)
    ;
  p++;
  long long res = 0;
  n += 2;
  for (int i = 60; i >= 0; i--) {
    if ((1LL << (i)) & n) res += c[i][p];
    p -= ((1LL << (i)) & n) > 0;
    if (p < 0) break;
  }
  if (t == 1 && n == 1) res++;
  if (t == 1 && n != 1) res--;
  cout << res << endl;
  return 0;
}

```