#include <bits/stdc++.h>
using namespace std;
bool check(long long n, long long pos) { return (bool)(n & (1ll << pos)); }
int main() {
  long long s, x, A, cnt = 0, ans;
  scanf("%lld", &s), scanf("%lld", &x);
  A = (s - x) / 2;
  if ((s - x) & 1) {
    A++;
  }
  for (long long i = 50; i >= 0; i--) {
    if (check(x, i) and check(A, i)) {
      printf("0\n");
      exit(0);
    }
    if (check(A, i)) {
      continue;
    } else if (check(x, i)) {
      cnt++;
    }
  }
  ans = 1ll << cnt;
  if (s == x) {
    ans -= 2;
    printf("%lld\n", ans);
  } else {
    printf("%lld\n", ans);
  }
  exit(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;
bool check(long long n, long long pos) { return (bool)(n & (1ll << pos)); }
int main() {
  long long s, x, A, cnt = 0, ans;
  scanf("%lld", &s), scanf("%lld", &x);
  A = (s - x) / 2;
  if ((s - x) & 1) {
    A++;
  }
  for (long long i = 50; i >= 0; i--) {
    if (check(x, i) and check(A, i)) {
      printf("0\n");
      exit(0);
    }
    if (check(A, i)) {
      continue;
    } else if (check(x, i)) {
      cnt++;
    }
  }
  ans = 1ll << cnt;
  if (s == x) {
    ans -= 2;
    printf("%lld\n", ans);
  } else {
    printf("%lld\n", ans);
  }
  exit(0);
}

```