#include <bits/stdc++.h>
int main(void) {
  long long i, j, p, q;
  while (~scanf("%I64d%I64d", &p, &q)) {
    long long ans = (p - q);
    long long dd = ans / 2;
    if (p < q || ans % 2 || dd & q) {
      printf("0\n");
      continue;
    }
    long long KK = 1;
    while (q) {
      if (q % 2) KK *= 2;
      q /= 2;
    }
    if (ans == 0) KK -= 2;
    printf("%I64d\n", KK);
  }
  return 0;
}

### Prompt

Your task is to create a Cpp solution to the following problem:

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>
int main(void) {
  long long i, j, p, q;
  while (~scanf("%I64d%I64d", &p, &q)) {
    long long ans = (p - q);
    long long dd = ans / 2;
    if (p < q || ans % 2 || dd & q) {
      printf("0\n");
      continue;
    }
    long long KK = 1;
    while (q) {
      if (q % 2) KK *= 2;
      q /= 2;
    }
    if (ans == 0) KK -= 2;
    printf("%I64d\n", KK);
  }
  return 0;
}

```