#include <bits/stdc++.h>
using namespace std;
const int D = 60;
long long sum, xsum;
int main() {
  scanf("%I64d%I64d", &sum, &xsum);
  long long ans = 1, res = 0;
  for (int i = 0; i < D; ++i) {
    if (xsum >> i & 1)
      ans <<= 1, sum -= 1ll << i;
    else
      res |= 1ll << i;
  }
  res <<= 1;
  if ((res | sum) != res) return puts("0"), 0;
  if (!sum) ans -= 2;
  printf("%I64d\n", 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 D = 60;
long long sum, xsum;
int main() {
  scanf("%I64d%I64d", &sum, &xsum);
  long long ans = 1, res = 0;
  for (int i = 0; i < D; ++i) {
    if (xsum >> i & 1)
      ans <<= 1, sum -= 1ll << i;
    else
      res |= 1ll << i;
  }
  res <<= 1;
  if ((res | sum) != res) return puts("0"), 0;
  if (!sum) ans -= 2;
  printf("%I64d\n", ans);
  return 0;
}

```