#include <bits/stdc++.h>
using namespace std;
int main() {
  long long s, x;
  cin >> s >> x;
  long long ans = 1;
  for (long long k = 1; k <= x; k *= 2LL) {
    if ((x & k) > 0LL) {
      ans *= 2LL;
      s -= k;
    }
  }
  if (s < 0 || s % 2LL == 1LL) {
    printf("0");
    return 0;
  }
  s /= 2;
  if ((s & x) > 0) {
    printf("0");
    return 0;
  } else if (s == 0)
    ans -= 2;
  cout << ans;
}

### Prompt

In CPP, your task is to solve 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 s, x;
  cin >> s >> x;
  long long ans = 1;
  for (long long k = 1; k <= x; k *= 2LL) {
    if ((x & k) > 0LL) {
      ans *= 2LL;
      s -= k;
    }
  }
  if (s < 0 || s % 2LL == 1LL) {
    printf("0");
    return 0;
  }
  s /= 2;
  if ((s & x) > 0) {
    printf("0");
    return 0;
  } else if (s == 0)
    ans -= 2;
  cout << ans;
}

```