#include <bits/stdc++.h>
using namespace std;
const int N = 1 << 6;
const int inf = 50;
long long s, x;
long long dp[N][2];
long long solve(int idx, int carry) {
  if (idx >= inf) return !carry;
  long long &ret = dp[idx][carry];
  if (ret != -1) return ret;
  ret = 0;
  for (int i = 0; i < 2; i++) {
    for (int j = 0; j < 2; j++) {
      if ((i + j + carry) % 2 != (s >> idx & 1)) continue;
      if ((i ^ j) != (x >> idx & 1)) continue;
      ret += solve(idx + 1, (i + j + carry) / 2);
    }
  }
  return ret;
}
int main() {
  memset(dp, -1, sizeof(dp));
  cin >> s >> x;
  long long ans = solve(0, 0);
  if (s == x) ans -= 2;
  cout << ans << endl;
  return 0;
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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 = 1 << 6;
const int inf = 50;
long long s, x;
long long dp[N][2];
long long solve(int idx, int carry) {
  if (idx >= inf) return !carry;
  long long &ret = dp[idx][carry];
  if (ret != -1) return ret;
  ret = 0;
  for (int i = 0; i < 2; i++) {
    for (int j = 0; j < 2; j++) {
      if ((i + j + carry) % 2 != (s >> idx & 1)) continue;
      if ((i ^ j) != (x >> idx & 1)) continue;
      ret += solve(idx + 1, (i + j + carry) / 2);
    }
  }
  return ret;
}
int main() {
  memset(dp, -1, sizeof(dp));
  cin >> s >> x;
  long long ans = solve(0, 0);
  if (s == x) ans -= 2;
  cout << ans << endl;
  return 0;
}

```