#include <bits/stdc++.h>
using namespace std;
const int INF = (int)1e9;
const int mod = (int)1e9 + 7;
const long long LLINF = (long long)1e18;
const int LOG = 17;
const int N = (int)2e5 + 5;
const double PI = 3.141592653589793;
const double eps = 0.000005;
void operator delete(void* a) {}
void operator delete[](void* a) {}
long long s, x;
int main() {
  cin >> s >> x;
  if (s < 0 || (s - x) & 1) {
    return puts("0"), 0;
  }
  s = (s - x) >> 1;
  long long ans = 1ll;
  for (int i = 0; i < 63; ++i) {
    if (((x >> i) & 1) == 1) {
      if (((s >> i) & 1) == 1) {
        return puts("0"), 0;
      }
      ans *= 2;
    }
  }
  if (!s) {
    ans -= 2;
  }
  cout << ans;
  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;
const int INF = (int)1e9;
const int mod = (int)1e9 + 7;
const long long LLINF = (long long)1e18;
const int LOG = 17;
const int N = (int)2e5 + 5;
const double PI = 3.141592653589793;
const double eps = 0.000005;
void operator delete(void* a) {}
void operator delete[](void* a) {}
long long s, x;
int main() {
  cin >> s >> x;
  if (s < 0 || (s - x) & 1) {
    return puts("0"), 0;
  }
  s = (s - x) >> 1;
  long long ans = 1ll;
  for (int i = 0; i < 63; ++i) {
    if (((x >> i) & 1) == 1) {
      if (((s >> i) & 1) == 1) {
        return puts("0"), 0;
      }
      ans *= 2;
    }
  }
  if (!s) {
    ans -= 2;
  }
  cout << ans;
  return 0;
}

```