#include <bits/stdc++.h>
using namespace std;
int main() {
  long long int sum, XOR, AND;
  scanf("%I64d%I64d", &sum, &XOR);
  if ((sum - XOR) % 2) {
    printf("0\n");
    return 0;
  }
  AND = (sum - XOR) / 2;
  long long int ans = 1;
  for (long long int digit = 0; digit <= 60; digit++) {
    int a = 0, x = 0;
    if (AND & (1LL << digit)) a = 1;
    if (XOR & (1LL << digit)) x = 1;
    if (a) {
      if (x) ans = 0;
    } else {
      if (x) ans = ans * 2LL;
    }
  }
  if (XOR == sum) ans -= 2;
  printf("%I64d\n", ans);
  return 0;
}

### Prompt

Please create a solution in Cpp 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>
using namespace std;
int main() {
  long long int sum, XOR, AND;
  scanf("%I64d%I64d", &sum, &XOR);
  if ((sum - XOR) % 2) {
    printf("0\n");
    return 0;
  }
  AND = (sum - XOR) / 2;
  long long int ans = 1;
  for (long long int digit = 0; digit <= 60; digit++) {
    int a = 0, x = 0;
    if (AND & (1LL << digit)) a = 1;
    if (XOR & (1LL << digit)) x = 1;
    if (a) {
      if (x) ans = 0;
    } else {
      if (x) ans = ans * 2LL;
    }
  }
  if (XOR == sum) ans -= 2;
  printf("%I64d\n", ans);
  return 0;
}

```