#include <bits/stdc++.h>
using namespace std;
long long solve(long long a, long long b, int add) {
  if (a == 0) {
    return b == 0;
  }
  if (b % 2) {
    if ((add + 1) % 2 != a % 2) return 0;
    return 2 * solve(a / 2, b / 2, (add + 1) / 2);
  } else {
    if (add != a % 2) return 0;
    long long ans = solve(a / 2, b / 2, 0);
    if (a > 1) ans += solve(a / 2, b / 2, 1);
    return ans;
  }
}
int main() {
  long long s, x;
  cin >> s >> x;
  long long ans = solve(s, x, 0);
  if (s == x) ans -= 2;
  cout << ans << endl;
  return 0;
}

### Prompt

In cpp, your task is to solve 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 solve(long long a, long long b, int add) {
  if (a == 0) {
    return b == 0;
  }
  if (b % 2) {
    if ((add + 1) % 2 != a % 2) return 0;
    return 2 * solve(a / 2, b / 2, (add + 1) / 2);
  } else {
    if (add != a % 2) return 0;
    long long ans = solve(a / 2, b / 2, 0);
    if (a > 1) ans += solve(a / 2, b / 2, 1);
    return ans;
  }
}
int main() {
  long long s, x;
  cin >> s >> x;
  long long ans = solve(s, x, 0);
  if (s == x) ans -= 2;
  cout << ans << endl;
  return 0;
}

```