#include <bits/stdc++.h>
long long a, b;
long long B[100], Z[100];
int main() {
  scanf("%I64d%I64d", &a, &b);
  B[0] = 1;
  for (int i = 0; i < 60; i++) {
    if ((a & (1ll << i)) == 0 && (b & (1ll << i)) == 0) {
      B[i + 1] = B[i];
      Z[i + 1] = B[i];
    }
    if ((a & (1ll << i)) == 0 && (b & (1ll << i)) != 0) {
      B[i + 1] = 0;
      Z[i + 1] = Z[i] * 2;
    }
    if ((a & (1ll << i)) != 0 && (b & (1ll << i)) == 0) {
      B[i + 1] = Z[i];
      Z[i + 1] = Z[i];
    }
    if ((a & (1ll << i)) != 0 && (b & (1ll << i)) != 0) {
      B[i + 1] = B[i] * 2;
      Z[i + 1] = 0;
    }
  }
  if (a == b) B[60] -= 2;
  printf("%I64d", B[60]);
}

### 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>
long long a, b;
long long B[100], Z[100];
int main() {
  scanf("%I64d%I64d", &a, &b);
  B[0] = 1;
  for (int i = 0; i < 60; i++) {
    if ((a & (1ll << i)) == 0 && (b & (1ll << i)) == 0) {
      B[i + 1] = B[i];
      Z[i + 1] = B[i];
    }
    if ((a & (1ll << i)) == 0 && (b & (1ll << i)) != 0) {
      B[i + 1] = 0;
      Z[i + 1] = Z[i] * 2;
    }
    if ((a & (1ll << i)) != 0 && (b & (1ll << i)) == 0) {
      B[i + 1] = Z[i];
      Z[i + 1] = Z[i];
    }
    if ((a & (1ll << i)) != 0 && (b & (1ll << i)) != 0) {
      B[i + 1] = B[i] * 2;
      Z[i + 1] = 0;
    }
  }
  if (a == b) B[60] -= 2;
  printf("%I64d", B[60]);
}

```