#include <bits/stdc++.h>
using namespace std;
long long s, x, cnt, asum, res;
int main() {
  while (~scanf("%I64d%I64d", &s, &x)) {
    cnt = res = 0, asum = (s - x) >> 1;
    if ((s - x) % 2 == 0 && s - x >= 0) {
      bool flag = true;
      if (s == x) res -= 2;
      while (x) {
        if (x & 1) {
          cnt++;
          if (asum & 1) flag = false;
        }
        if (!flag) break;
        x >>= 1, asum >>= 1;
      }
      if (!flag)
        res = 0;
      else
        res += 1LL << cnt;
    }
    printf("%I64d\n", res);
  }
  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;
long long s, x, cnt, asum, res;
int main() {
  while (~scanf("%I64d%I64d", &s, &x)) {
    cnt = res = 0, asum = (s - x) >> 1;
    if ((s - x) % 2 == 0 && s - x >= 0) {
      bool flag = true;
      if (s == x) res -= 2;
      while (x) {
        if (x & 1) {
          cnt++;
          if (asum & 1) flag = false;
        }
        if (!flag) break;
        x >>= 1, asum >>= 1;
      }
      if (!flag)
        res = 0;
      else
        res += 1LL << cnt;
    }
    printf("%I64d\n", res);
  }
  return 0;
}

```