#include <bits/stdc++.h>
using namespace std;
long long s, x, a;
long long otv, i;
int main() {
  scanf("%I64d %I64d", &s, &x);
  if ((s - x) & 1 == 1) {
    printf("0\n");
    return 0;
  }
  a = (s - x) >> 1;
  i = 0;
  while (true) {
    if ((x >> i) == 0) break;
    if ((x >> i) & 1 == 1 && (a >> i) & 1 == 1) {
      printf("0\n");
      return 0;
    }
    if ((x >> i) & 1 == 1) otv++;
    i++;
  }
  otv = 1ll << otv;
  if (a == 0) otv -= 2;
  otv = max(0ll, otv);
  printf("%I64d\n", otv);
  return 0;
}

### Prompt

Your challenge is to write 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>
using namespace std;
long long s, x, a;
long long otv, i;
int main() {
  scanf("%I64d %I64d", &s, &x);
  if ((s - x) & 1 == 1) {
    printf("0\n");
    return 0;
  }
  a = (s - x) >> 1;
  i = 0;
  while (true) {
    if ((x >> i) == 0) break;
    if ((x >> i) & 1 == 1 && (a >> i) & 1 == 1) {
      printf("0\n");
      return 0;
    }
    if ((x >> i) & 1 == 1) otv++;
    i++;
  }
  otv = 1ll << otv;
  if (a == 0) otv -= 2;
  otv = max(0ll, otv);
  printf("%I64d\n", otv);
  return 0;
}

```