#include <bits/stdc++.h>
using namespace std;
int main() {
  long long s, x;
  scanf("%lld %lld", &s, &x);
  if (s % 2 != x % 2) {
    puts("0");
    return 0;
  }
  int reg = 0;
  long long ans = 1;
  for (int i = 0; i < 50; i++) {
    int snxt = ((s >> (i + 1)) & 1), xnxt = ((x >> (i + 1)) & 1);
    int scur = ((s >> i) & 1), xcur = ((x >> i) & 1);
    int need = (snxt != xnxt);
    int fac = 0;
    for (int j = 0; j < 2; j++) {
      for (int k = 0; k < 2; k++) {
        int a = reg + j + k;
        if (need == (a >= 2) && a % 2 == scur && (j + k) % 2 == xcur) {
          fac++;
        }
      }
    }
    ans *= fac;
    reg = need;
  }
  if (s == x) {
    ans -= 2;
  }
  printf("%lld\n", ans);
}

### Prompt

Please formulate 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>
using namespace std;
int main() {
  long long s, x;
  scanf("%lld %lld", &s, &x);
  if (s % 2 != x % 2) {
    puts("0");
    return 0;
  }
  int reg = 0;
  long long ans = 1;
  for (int i = 0; i < 50; i++) {
    int snxt = ((s >> (i + 1)) & 1), xnxt = ((x >> (i + 1)) & 1);
    int scur = ((s >> i) & 1), xcur = ((x >> i) & 1);
    int need = (snxt != xnxt);
    int fac = 0;
    for (int j = 0; j < 2; j++) {
      for (int k = 0; k < 2; k++) {
        int a = reg + j + k;
        if (need == (a >= 2) && a % 2 == scur && (j + k) % 2 == xcur) {
          fac++;
        }
      }
    }
    ans *= fac;
    reg = need;
  }
  if (s == x) {
    ans -= 2;
  }
  printf("%lld\n", ans);
}

```