#include <bits/stdc++.h>
using namespace std;
int main() {
  long long s, x;
  while (cin >> s >> x) {
    long long dp[41][2];
    memset(dp, 0, sizeof dp);
    dp[0][0] = 1;
    for (int i = 0; i < 40; i++) {
      if (x & (1LL << i)) {
        if (s & (1LL << i)) dp[i + 1][0] += 2 * dp[i][0];
        if (!(s & (1LL << i))) dp[i + 1][1] += 2 * dp[i][1];
      } else {
        if (!(s & (1LL << i)))
          dp[i + 1][0] += dp[i][0], dp[i + 1][1] += dp[i][0];
        if (s & (1LL << i)) dp[i + 1][0] += dp[i][1], dp[i + 1][1] += dp[i][1];
      }
    }
    cout << dp[40][0] - 2 * (s == x) << endl;
  }
  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;
int main() {
  long long s, x;
  while (cin >> s >> x) {
    long long dp[41][2];
    memset(dp, 0, sizeof dp);
    dp[0][0] = 1;
    for (int i = 0; i < 40; i++) {
      if (x & (1LL << i)) {
        if (s & (1LL << i)) dp[i + 1][0] += 2 * dp[i][0];
        if (!(s & (1LL << i))) dp[i + 1][1] += 2 * dp[i][1];
      } else {
        if (!(s & (1LL << i)))
          dp[i + 1][0] += dp[i][0], dp[i + 1][1] += dp[i][0];
        if (s & (1LL << i)) dp[i + 1][0] += dp[i][1], dp[i + 1][1] += dp[i][1];
      }
    }
    cout << dp[40][0] - 2 * (s == x) << endl;
  }
  return 0;
}

```