#include <bits/stdc++.h>
using namespace std;
long long s, x;
long long done[45][45][2][2][2];
long long calc(int xd, int sd, int c, bool b1nz, bool b2nz) {
  if (xd == 44 && !b1nz) return 0;
  if (xd == 44 && !b2nz) return 0;
  if (xd == 44) return c == 0 ? 1 : 0;
  long long &ans = done[xd][sd][c][b1nz][b2nz];
  if (ans != -1) return ans;
  ans = 0;
  for (int b1 = 0; b1 <= 1; b1++) {
    for (int b2 = 0; b2 <= 1; b2++) {
      if (((c + b1 + b2) % 2 == ((s >> sd) & 1)) &&
          ((b1 ^ b2) == ((x >> xd) & 1))) {
        ans = ans +
              calc(xd + 1, sd + 1, (c + b1 + b2) / 2, b1nz || b1, b2nz || b2);
      }
    }
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> s >> x;
  memset(done, -1, sizeof done);
  long long ans = calc(0, 0, 0, 0, 0);
  cout << ans;
  return 0;
}

### Prompt

Generate 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 s, x;
long long done[45][45][2][2][2];
long long calc(int xd, int sd, int c, bool b1nz, bool b2nz) {
  if (xd == 44 && !b1nz) return 0;
  if (xd == 44 && !b2nz) return 0;
  if (xd == 44) return c == 0 ? 1 : 0;
  long long &ans = done[xd][sd][c][b1nz][b2nz];
  if (ans != -1) return ans;
  ans = 0;
  for (int b1 = 0; b1 <= 1; b1++) {
    for (int b2 = 0; b2 <= 1; b2++) {
      if (((c + b1 + b2) % 2 == ((s >> sd) & 1)) &&
          ((b1 ^ b2) == ((x >> xd) & 1))) {
        ans = ans +
              calc(xd + 1, sd + 1, (c + b1 + b2) / 2, b1nz || b1, b2nz || b2);
      }
    }
  }
  return ans;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> s >> x;
  memset(done, -1, sizeof done);
  long long ans = calc(0, 0, 0, 0, 0);
  cout << ans;
  return 0;
}

```