#include <bits/stdc++.h>
using namespace std;
const int N = 1e6 + 6, inf = 1e9 + 7;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long int n, s;
  cin >> n >> s;
  long long int d = n - s;
  if (d & 1) cout << "0\n", exit(0);
  d /= 2;
  long long int res = 1;
  for (int i = 0; i < 60; i++) {
    if (s & (1LL << i)) {
      res *= 2LL;
      if (d & (1LL << i)) cout << "0\n", exit(0);
    }
  }
  if (n == s) res -= 2;
  cout << res << "\n";
  return 0;
}

### Prompt

In CPP, your task is to solve 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 = 1e6 + 6, inf = 1e9 + 7;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long int n, s;
  cin >> n >> s;
  long long int d = n - s;
  if (d & 1) cout << "0\n", exit(0);
  d /= 2;
  long long int res = 1;
  for (int i = 0; i < 60; i++) {
    if (s & (1LL << i)) {
      res *= 2LL;
      if (d & (1LL << i)) cout << "0\n", exit(0);
    }
  }
  if (n == s) res -= 2;
  cout << res << "\n";
  return 0;
}

```