#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
void solve(int test) {
  long long S, X;
  cin >> S >> X;
  if ((S - X) % 2 || S < X) {
    cout << 0 << "\n";
    return;
  }
  long long A = (S - X) / 2;
  long long ans = 1;
  for (int i = 60; i >= 0; i--) {
    int x = (A >> i) & 1;
    int y = (X >> i) & 1;
    if (x == 0 && y == 1)
      ans *= 2;
    else if (x == 1 && y == 1)
      ans = 0;
  }
  if (A == 0) ans -= 2;
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  for (int test = 1; test <= t; test++) {
    solve(test);
  }
  return 0;
}

### Prompt

Please create a solution in cpp 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;
const int N = 2e5 + 5;
void solve(int test) {
  long long S, X;
  cin >> S >> X;
  if ((S - X) % 2 || S < X) {
    cout << 0 << "\n";
    return;
  }
  long long A = (S - X) / 2;
  long long ans = 1;
  for (int i = 60; i >= 0; i--) {
    int x = (A >> i) & 1;
    int y = (X >> i) & 1;
    if (x == 0 && y == 1)
      ans *= 2;
    else if (x == 1 && y == 1)
      ans = 0;
  }
  if (A == 0) ans -= 2;
  cout << ans << "\n";
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  int t = 1;
  for (int test = 1; test <= t; test++) {
    solve(test);
  }
  return 0;
}

```