#include <bits/stdc++.h>
int main() {
  long long s, x;
  int p1, p2, i, judge = 0;
  scanf("%lld%lld", &s, &x);
  if ((s - x) % 2 != 0)
    printf("0\n");
  else {
    long long ans = 1;
    if (s == x) judge = 1;
    int flag = 1;
    s = (s - x) / 2;
    for (i = 0; i < 60; i++) {
      p1 = (s >> i) & 1;
      p2 = (x >> i) & 1;
      if (p1 == 1 && p2 == 1) {
        printf("0\n");
        flag = 0;
        break;
      }
      if (p2 == 1) ans *= 2;
    }
    if (flag) {
      if (judge) ans -= 2;
      printf("%lld\n", ans);
    }
  }
}

### Prompt

Please provide a Cpp coded solution to the problem described below:

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() {
  long long s, x;
  int p1, p2, i, judge = 0;
  scanf("%lld%lld", &s, &x);
  if ((s - x) % 2 != 0)
    printf("0\n");
  else {
    long long ans = 1;
    if (s == x) judge = 1;
    int flag = 1;
    s = (s - x) / 2;
    for (i = 0; i < 60; i++) {
      p1 = (s >> i) & 1;
      p2 = (x >> i) & 1;
      if (p1 == 1 && p2 == 1) {
        printf("0\n");
        flag = 0;
        break;
      }
      if (p2 == 1) ans *= 2;
    }
    if (flag) {
      if (judge) ans -= 2;
      printf("%lld\n", ans);
    }
  }
}

```