#include <bits/stdc++.h>
using namespace std;
long long a, b, c, tem, flag;
int main() {
  scanf("%I64d%I64d", &a, &b);
  c = a - b;
  if (c < 0 || c & 1) {
    puts("0");
    return 0;
  }
  c >>= 1;
  tem = 1;
  if (c == 0) flag = -2;
  while (b || c) {
    if (b & 1) {
      tem = (c & 1) ? 0 : (tem << 1);
    }
    b >>= 1;
    c >>= 1;
  }
  printf("%I64d\n", tem + flag);
  return 0;
}

### Prompt

Please provide a CPP coded solution 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;
long long a, b, c, tem, flag;
int main() {
  scanf("%I64d%I64d", &a, &b);
  c = a - b;
  if (c < 0 || c & 1) {
    puts("0");
    return 0;
  }
  c >>= 1;
  tem = 1;
  if (c == 0) flag = -2;
  while (b || c) {
    if (b & 1) {
      tem = (c & 1) ? 0 : (tem << 1);
    }
    b >>= 1;
    c >>= 1;
  }
  printf("%I64d\n", tem + flag);
  return 0;
}

```