#include <bits/stdc++.h>
using namespace std;
long long s, x;
int main() {
  while (scanf("%I64d%I64d", &s, &x) == 2) {
    long long a = (s - x) / 2;
    if (x > s || a * 2 + x != s || (x & a) != 0) {
      printf("0\n");
      continue;
    }
    long long bit = 0;
    while (x) {
      if (x & 1) bit++;
      x >>= 1;
    }
    long long res = 1ll << bit;
    if (a == 0) res -= 2;
    printf("%I64d\n", res);
  }
  return 0;
}

### Prompt

Please create a solution in CPP 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>
using namespace std;
long long s, x;
int main() {
  while (scanf("%I64d%I64d", &s, &x) == 2) {
    long long a = (s - x) / 2;
    if (x > s || a * 2 + x != s || (x & a) != 0) {
      printf("0\n");
      continue;
    }
    long long bit = 0;
    while (x) {
      if (x & 1) bit++;
      x >>= 1;
    }
    long long res = 1ll << bit;
    if (a == 0) res -= 2;
    printf("%I64d\n", res);
  }
  return 0;
}

```