#include <bits/stdc++.h>
using namespace std;
long long n, m, dp[40][2];
long long solve(int i, bool c) {
  bool s = n & (1LL << i);
  bool x = m & (1LL << i);
  if (c ^ s ^ x) return 0;
  if (1LL << i > m) return 1;
  if (dp[i][c] != -1) return dp[i][c];
  return dp[i][c] = solve(i + 1, x ? c : 1) + solve(i + 1, x ? c : 0);
}
int main() {
  cin >> n >> m;
  memset(dp, -1, sizeof(dp));
  cout << (solve(0, 0) - (n == m ? 2 : 0)) << endl;
  return 0;
}

### Prompt

Please formulate 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;
long long n, m, dp[40][2];
long long solve(int i, bool c) {
  bool s = n & (1LL << i);
  bool x = m & (1LL << i);
  if (c ^ s ^ x) return 0;
  if (1LL << i > m) return 1;
  if (dp[i][c] != -1) return dp[i][c];
  return dp[i][c] = solve(i + 1, x ? c : 1) + solve(i + 1, x ? c : 0);
}
int main() {
  cin >> n >> m;
  memset(dp, -1, sizeof(dp));
  cout << (solve(0, 0) - (n == m ? 2 : 0)) << endl;
  return 0;
}

```