#include <bits/stdc++.h>
using namespace std;
vector<bool> sum, op;
int main() {
  ios_base::sync_with_stdio(false);
  long long s, x, cs, cx;
  cin >> s >> x;
  cs = s;
  cx = x;
  long long ways_c = 0, ways = 1;
  while (x != 0 or s != 0) {
    int bone = s % 2, btwo = x % 2;
    if (bone and btwo) {
      ways = ways * 2;
      ways_c = 0;
    } else if (bone) {
      ways = ways_c;
      ways_c = ways_c;
    } else if (btwo) {
      ways = 0;
      ways_c = ways_c * 2;
    } else {
      ways = ways;
      ways_c = ways;
    }
    x /= 2;
    s /= 2;
  }
  if (cs == cx) ways -= 2;
  cout << ways << endl;
}

### 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;
vector<bool> sum, op;
int main() {
  ios_base::sync_with_stdio(false);
  long long s, x, cs, cx;
  cin >> s >> x;
  cs = s;
  cx = x;
  long long ways_c = 0, ways = 1;
  while (x != 0 or s != 0) {
    int bone = s % 2, btwo = x % 2;
    if (bone and btwo) {
      ways = ways * 2;
      ways_c = 0;
    } else if (bone) {
      ways = ways_c;
      ways_c = ways_c;
    } else if (btwo) {
      ways = 0;
      ways_c = ways_c * 2;
    } else {
      ways = ways;
      ways_c = ways;
    }
    x /= 2;
    s /= 2;
  }
  if (cs == cx) ways -= 2;
  cout << ways << endl;
}

```