#include <bits/stdc++.h>
int main() {
  long long x, n;
  scanf("%lld%lld", &x, &n);
  if (x < n || ((x - n) & 1) != 0) {
    puts("0");
    return 0;
  }
  x = (x - n) >> 1;
  bool all = n != 0, fuck = x == 0;
  int cnt = 0;
  for (; n != 0; n >>= 1, x >>= 1) {
    if ((n & 1) && !(x & 1)) {
      ++cnt;
    } else if (!(n & 1)) {
      all = false;
    } else {
      puts("0");
      return 0;
    }
  }
  long long ans = 1;
  for (; cnt != 0; --cnt) ans <<= 1;
  if (all || fuck) {
    ans -= 2;
  }
  printf("%lld", ans);
  return 0;
}

### Prompt

Generate a cpp solution 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>
int main() {
  long long x, n;
  scanf("%lld%lld", &x, &n);
  if (x < n || ((x - n) & 1) != 0) {
    puts("0");
    return 0;
  }
  x = (x - n) >> 1;
  bool all = n != 0, fuck = x == 0;
  int cnt = 0;
  for (; n != 0; n >>= 1, x >>= 1) {
    if ((n & 1) && !(x & 1)) {
      ++cnt;
    } else if (!(n & 1)) {
      all = false;
    } else {
      puts("0");
      return 0;
    }
  }
  long long ans = 1;
  for (; cnt != 0; --cnt) ans <<= 1;
  if (all || fuck) {
    ans -= 2;
  }
  printf("%lld", ans);
  return 0;
}

```