#include <bits/stdc++.h>
using namespace std;
long long sum, xr, ans = 1, i;
bool H[50];
int main() {
  scanf("%lld %lld", &sum, &xr);
  for (i = 0; i < 45; i++)
    if ((xr >> i) & 1LL) {
      ans *= 2LL;
      H[i] = 1;
    }
  if (xr > sum) return cout << 0, 0;
  sum -= xr;
  if (sum & 1LL) return cout << 0, 0;
  sum /= 2LL;
  for (i = 0; i < 45; i++)
    if ((sum >> i) & 1LL) {
      if (H[i]) return cout << 0, 0;
    }
  if (!sum) ans -= 2LL;
  printf("%lld", ans);
}

### 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 sum, xr, ans = 1, i;
bool H[50];
int main() {
  scanf("%lld %lld", &sum, &xr);
  for (i = 0; i < 45; i++)
    if ((xr >> i) & 1LL) {
      ans *= 2LL;
      H[i] = 1;
    }
  if (xr > sum) return cout << 0, 0;
  sum -= xr;
  if (sum & 1LL) return cout << 0, 0;
  sum /= 2LL;
  for (i = 0; i < 45; i++)
    if ((sum >> i) & 1LL) {
      if (H[i]) return cout << 0, 0;
    }
  if (!sum) ans -= 2LL;
  printf("%lld", ans);
}

```