#include <bits/stdc++.h>
using namespace std;
unsigned long long int dp[2][60];
int main() {
  unsigned long long int s, x;
  cin >> s >> x;
  dp[0][0] = 1;
  unsigned long long int ONE = 1;
  for (int i = 1; i < 50; i++) {
    if ((x & (ONE << (i - 1))) == 0) {
      if ((s & (ONE << (i - 1))) == 0) {
        dp[0][i] += dp[0][i - 1];
        dp[1][i] += dp[0][i - 1];
      } else {
        dp[0][i] += dp[1][i - 1];
        dp[1][i] += dp[1][i - 1];
      }
    } else {
      if ((s & (ONE << (i - 1))) == 0) {
        dp[0][i] = 0;
        dp[1][i] += 2 * dp[1][i - 1];
      } else {
        dp[0][i] += 2 * dp[0][i - 1];
        dp[1][i] = 0;
      }
    }
  }
  if (s == x) {
    dp[0][49] -= 2;
  }
  cout << dp[0][49] << endl;
  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;
unsigned long long int dp[2][60];
int main() {
  unsigned long long int s, x;
  cin >> s >> x;
  dp[0][0] = 1;
  unsigned long long int ONE = 1;
  for (int i = 1; i < 50; i++) {
    if ((x & (ONE << (i - 1))) == 0) {
      if ((s & (ONE << (i - 1))) == 0) {
        dp[0][i] += dp[0][i - 1];
        dp[1][i] += dp[0][i - 1];
      } else {
        dp[0][i] += dp[1][i - 1];
        dp[1][i] += dp[1][i - 1];
      }
    } else {
      if ((s & (ONE << (i - 1))) == 0) {
        dp[0][i] = 0;
        dp[1][i] += 2 * dp[1][i - 1];
      } else {
        dp[0][i] += 2 * dp[0][i - 1];
        dp[1][i] = 0;
      }
    }
  }
  if (s == x) {
    dp[0][49] -= 2;
  }
  cout << dp[0][49] << endl;
  return 0;
}

```