#include <bits/stdc++.h>
int main() {
  long long unsigned s, x, a, b, r;
  unsigned o = 1;
  scanf("%I64u%I64u", &s, &x);
  for (a = x, b = 1; b <= s; b <<= 1)
    ;
  for (; b >>= 1;)
    if (!(b & x)) {
      if (a + (b << 1) <= s) {
        a += b << 1;
        o = 0;
      }
      if (a == s) break;
    }
  if (a != s) {
    printf("0\n");
    return 0;
  }
  for (r = b = 1; b <= x; b <<= 1)
    if (b & x) r <<= 1;
  if (o) r -= 2;
  printf("%I64u\n", r);
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 unsigned s, x, a, b, r;
  unsigned o = 1;
  scanf("%I64u%I64u", &s, &x);
  for (a = x, b = 1; b <= s; b <<= 1)
    ;
  for (; b >>= 1;)
    if (!(b & x)) {
      if (a + (b << 1) <= s) {
        a += b << 1;
        o = 0;
      }
      if (a == s) break;
    }
  if (a != s) {
    printf("0\n");
    return 0;
  }
  for (r = b = 1; b <= x; b <<= 1)
    if (b & x) r <<= 1;
  if (o) r -= 2;
  printf("%I64u\n", r);
  return 0;
}

```