#include <bits/stdc++.h>
using namespace std;
long long s, x;
int main() {
  int t;
  while (scanf("%lld%lld", &s, &x) != EOF) {
    long long ans = 0;
    s -= x;
    if (s % 2 == 1) {
      printf("0\n");
      continue;
    }
    s /= 2;
    int flag = 0, cnt = 0;
    if (!s) cnt = 2;
    for (int i = 0; i < 64; i++) {
      int a1 = s & 1;
      int a2 = x & 1;
      if (a1 == 1 && a2 == 1) {
        flag = 1;
        break;
      }
      s >>= 1;
      x >>= 1;
      if (a2 == 0) continue;
      ans++;
    }
    if (flag)
      printf("0\n");
    else
      printf("%lld\n", (1ll << ans) - cnt);
  }
  return 0;
}

### Prompt

Develop a solution in Cpp 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>
using namespace std;
long long s, x;
int main() {
  int t;
  while (scanf("%lld%lld", &s, &x) != EOF) {
    long long ans = 0;
    s -= x;
    if (s % 2 == 1) {
      printf("0\n");
      continue;
    }
    s /= 2;
    int flag = 0, cnt = 0;
    if (!s) cnt = 2;
    for (int i = 0; i < 64; i++) {
      int a1 = s & 1;
      int a2 = x & 1;
      if (a1 == 1 && a2 == 1) {
        flag = 1;
        break;
      }
      s >>= 1;
      x >>= 1;
      if (a2 == 0) continue;
      ans++;
    }
    if (flag)
      printf("0\n");
    else
      printf("%lld\n", (1ll << ans) - cnt);
  }
  return 0;
}

```