#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
const int MOD = 1e9 + 9;
const int MOD2 = 1e9 + 7;
int main(int argc, char const *argv[]) {
  cin.sync_with_stdio(0);
  cin.tie(0);
  long long int ans = 1;
  long long int s, x;
  cin >> s >> x;
  long long int os = s;
  if (s < x) {
    cout << "0\n";
    return 0;
  }
  if ((s - x) % 2 != 0) {
    cout << "0\n";
    return 0;
  }
  s = (s - x) / 2;
  for (int i = 0; i < 42; i++) {
    if ((1LL << i) > 1e12) break;
    if ((1LL << i) & x) {
      long long int v = (1LL << (i));
      if (v & s) {
        cout << 0 << '\n';
        return 0;
      }
      ans *= 2;
    }
  }
  if (os == x) ans -= 2;
  cout << max(0LL, ans) << '\n';
  return 0;
}

### Prompt

Create a solution in Cpp for 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;
const int N = 2e5 + 5;
const int MOD = 1e9 + 9;
const int MOD2 = 1e9 + 7;
int main(int argc, char const *argv[]) {
  cin.sync_with_stdio(0);
  cin.tie(0);
  long long int ans = 1;
  long long int s, x;
  cin >> s >> x;
  long long int os = s;
  if (s < x) {
    cout << "0\n";
    return 0;
  }
  if ((s - x) % 2 != 0) {
    cout << "0\n";
    return 0;
  }
  s = (s - x) / 2;
  for (int i = 0; i < 42; i++) {
    if ((1LL << i) > 1e12) break;
    if ((1LL << i) & x) {
      long long int v = (1LL << (i));
      if (v & s) {
        cout << 0 << '\n';
        return 0;
      }
      ans *= 2;
    }
  }
  if (os == x) ans -= 2;
  cout << max(0LL, ans) << '\n';
  return 0;
}

```