#include <bits/stdc++.h>
using namespace std;
set<long long> S;
long long s, x;
int ret;
int main() {
  ios_base::sync_with_stdio(false);
  cin >> s >> x;
  if (s == x) ret = 1;
  long long ans = 1;
  for (int i = 0; i <= 50; ++i) {
    if ((x >> i) & 1) {
      s -= (1LL << i);
      ans = ans * 2;
    } else {
      S.insert(i + 1);
    }
  }
  if (s < 0) {
    cout << 0;
    return 0;
  }
  for (int i = 0; i <= 50; ++i) {
    if ((s >> i) & 1) {
      if (S.find(i) == S.end()) {
        cout << 0;
        return 0;
      }
    }
  }
  ans = ans - 2 * ret;
  cout << ans;
}

### 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;
set<long long> S;
long long s, x;
int ret;
int main() {
  ios_base::sync_with_stdio(false);
  cin >> s >> x;
  if (s == x) ret = 1;
  long long ans = 1;
  for (int i = 0; i <= 50; ++i) {
    if ((x >> i) & 1) {
      s -= (1LL << i);
      ans = ans * 2;
    } else {
      S.insert(i + 1);
    }
  }
  if (s < 0) {
    cout << 0;
    return 0;
  }
  for (int i = 0; i <= 50; ++i) {
    if ((s >> i) & 1) {
      if (S.find(i) == S.end()) {
        cout << 0;
        return 0;
      }
    }
  }
  ans = ans - 2 * ret;
  cout << ans;
}

```