#include <bits/stdc++.h>
using namespace std;
int main() {
  long long s, x;
  while (~scanf("%I64d%I64d", &s, &x)) {
    if (s < x || (s - x) % 2) {
      printf("0\n");
      continue;
    }
    long long ans = 1;
    for (int i = 0; i < 63; i++)
      if ((1LL << i) & x) ans *= 2;
    if (s == x && x)
      ans -= 2;
    else if (s == x && !x)
      ans = 0;
    for (int i = 0; i < 63; i++)
      if ((1LL << i) & x)
        if ((1LL << i) & ((s - x) / 2)) ans = 0;
    printf("%I64d\n", ans);
  }
  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;
int main() {
  long long s, x;
  while (~scanf("%I64d%I64d", &s, &x)) {
    if (s < x || (s - x) % 2) {
      printf("0\n");
      continue;
    }
    long long ans = 1;
    for (int i = 0; i < 63; i++)
      if ((1LL << i) & x) ans *= 2;
    if (s == x && x)
      ans -= 2;
    else if (s == x && !x)
      ans = 0;
    for (int i = 0; i < 63; i++)
      if ((1LL << i) & x)
        if ((1LL << i) & ((s - x) / 2)) ans = 0;
    printf("%I64d\n", ans);
  }
  return 0;
}

```