#include <bits/stdc++.h>
using namespace std;
long long n, m;
int main() {
  scanf("%lld %lld", &n, &m);
  if (n < m || (n - m) % 2 || ((n - m) / 2) & m) {
    printf("0");
  } else {
    int sub = !((n - m) / 2) ? 2 : 0;
    long long ans = 1;
    while (m) {
      ans *= m % 2 ? 2 : 1;
      m /= 2;
    }
    printf("%lld", ans - sub);
  }
}

### Prompt

Construct a CPP code solution to the problem outlined:

Two positive integers a and b have a sum of s and a bitwise XOR of x. How many possible values are there for the ordered pair (a, b)?

Input

The first line of the input contains two integers s and x (2 ≤ s ≤ 1012, 0 ≤ x ≤ 1012), the sum and bitwise xor of the pair of positive integers, respectively.

Output

Print a single integer, the number of solutions to the given conditions. If no solutions exist, print 0.

Examples

Input

9 5


Output

4


Input

3 3


Output

2


Input

5 2


Output

0

Note

In the first sample, we have the following solutions: (2, 7), (3, 6), (6, 3), (7, 2).

In the second sample, the only solutions are (1, 2) and (2, 1).

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
long long n, m;
int main() {
  scanf("%lld %lld", &n, &m);
  if (n < m || (n - m) % 2 || ((n - m) / 2) & m) {
    printf("0");
  } else {
    int sub = !((n - m) / 2) ? 2 : 0;
    long long ans = 1;
    while (m) {
      ans *= m % 2 ? 2 : 1;
      m /= 2;
    }
    printf("%lld", ans - sub);
  }
}

```