#include <bits/stdc++.h>
using namespace std;
long long s, s1, x, sol = 1;
int main() {
  long long i, j;
  scanf("%I64d %I64d", &s, &x);
  for (i = 1; i <= x; i <<= 1)
    if (i & x) sol <<= 1;
  if (s == x) sol -= 2;
  s1 = s;
  s -= x;
  for (i = 1; i <= s; i <<= 1)
    if ((x & i) == 0) {
      if ((s & (i << 1))) s -= (i << 1);
    }
  if (s)
    printf("0\n");
  else
    printf("%I64d\n", sol);
  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;
long long s, s1, x, sol = 1;
int main() {
  long long i, j;
  scanf("%I64d %I64d", &s, &x);
  for (i = 1; i <= x; i <<= 1)
    if (i & x) sol <<= 1;
  if (s == x) sol -= 2;
  s1 = s;
  s -= x;
  for (i = 1; i <= s; i <<= 1)
    if ((x & i) == 0) {
      if ((s & (i << 1))) s -= (i << 1);
    }
  if (s)
    printf("0\n");
  else
    printf("%I64d\n", sol);
  return 0;
}

```