#include <bits/stdc++.h>
using namespace std;
long long f[50 + 5][2];
long long ss, xx;
void solve() {
  f[0][0] = 1;
  for (int i = 0; i < 50; i++)
    for (int carry = 0; carry < 2; carry++)
      for (int x = 0; x < 2; x++)
        for (int y = 0; y < 2; y++) {
          if (f[i][carry] == 0) continue;
          int bitIDx = ((xx & 1ll << (i)) > 0);
          int bitIDs = ((ss & 1ll << (i)) > 0);
          if ((x ^ y) != bitIDx) continue;
          if ((x ^ y ^ carry) != bitIDs) continue;
          f[i + 1][(x + y + carry) / 2] += f[i][carry];
        }
  cout << f[50][0] - (xx == ss) * 2;
}
int main() {
  cin >> ss >> xx;
  solve();
}

### Prompt

Your task is to create 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 f[50 + 5][2];
long long ss, xx;
void solve() {
  f[0][0] = 1;
  for (int i = 0; i < 50; i++)
    for (int carry = 0; carry < 2; carry++)
      for (int x = 0; x < 2; x++)
        for (int y = 0; y < 2; y++) {
          if (f[i][carry] == 0) continue;
          int bitIDx = ((xx & 1ll << (i)) > 0);
          int bitIDs = ((ss & 1ll << (i)) > 0);
          if ((x ^ y) != bitIDx) continue;
          if ((x ^ y ^ carry) != bitIDs) continue;
          f[i + 1][(x + y + carry) / 2] += f[i][carry];
        }
  cout << f[50][0] - (xx == ss) * 2;
}
int main() {
  cin >> ss >> xx;
  solve();
}

```