#include <bits/stdc++.h>
using namespace std;
long long a, b, x, y;
long long A[45];
long long B[45];
int ans, w;
int main() {
  scanf("%I64d %I64d", &a, &b);
  A[0] = 1ll;
  for (int i = (1); i < (40); i++) A[i] = A[i - 1] * 2ll;
  x = b;
  for (int i = (39); i >= (0); i--) {
    if (x >= A[i]) {
      x -= A[i];
      ans++;
    } else if (A[i] * 2ll <= a - b) {
      B[w++] = A[i] * 2ll;
    }
  }
  x = a - b;
  for (int i = (0); i < (w); i++)
    if (x >= B[i]) x -= B[i];
  if (a == b)
    printf("%I64d\n", A[ans] - 2ll);
  else if (x != 0ll)
    printf("0\n");
  else
    printf("%I64d\n", A[ans]);
  return 0;
}

### Prompt

Please formulate 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 a, b, x, y;
long long A[45];
long long B[45];
int ans, w;
int main() {
  scanf("%I64d %I64d", &a, &b);
  A[0] = 1ll;
  for (int i = (1); i < (40); i++) A[i] = A[i - 1] * 2ll;
  x = b;
  for (int i = (39); i >= (0); i--) {
    if (x >= A[i]) {
      x -= A[i];
      ans++;
    } else if (A[i] * 2ll <= a - b) {
      B[w++] = A[i] * 2ll;
    }
  }
  x = a - b;
  for (int i = (0); i < (w); i++)
    if (x >= B[i]) x -= B[i];
  if (a == b)
    printf("%I64d\n", A[ans] - 2ll);
  else if (x != 0ll)
    printf("0\n");
  else
    printf("%I64d\n", A[ans]);
  return 0;
}

```