#include <bits/stdc++.h>
using namespace std;
map<pair<pair<long long int, long long int>, int>, long long int> memo;
long long int dp(long long int s, long long int x, int traigoUno = 0) {
  if (memo.find({{s, x}, traigoUno}) != memo.end()) {
    return memo[{{s, x}, traigoUno}];
  }
  if (traigoUno == 0 && s == 0 && x == 0) return 1;
  long long int res = 0;
  for (int i = (0); i < (int)(2); i++) {
    int j = (x % 2) ^ i;
    if ((i + j + traigoUno) % 2 != s % 2) {
      continue;
    }
    res += dp(s / 2, x / 2, min(1, max(0, j + i + traigoUno - 1)));
  }
  memo[{{s, x}, traigoUno}] = res;
  return res;
}
int main() {
  long long int s, x;
  cin >> s >> x;
  cout << dp(s, x) - 2 * (s == x) << endl;
}

### Prompt

Create a solution in cpp for 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;
map<pair<pair<long long int, long long int>, int>, long long int> memo;
long long int dp(long long int s, long long int x, int traigoUno = 0) {
  if (memo.find({{s, x}, traigoUno}) != memo.end()) {
    return memo[{{s, x}, traigoUno}];
  }
  if (traigoUno == 0 && s == 0 && x == 0) return 1;
  long long int res = 0;
  for (int i = (0); i < (int)(2); i++) {
    int j = (x % 2) ^ i;
    if ((i + j + traigoUno) % 2 != s % 2) {
      continue;
    }
    res += dp(s / 2, x / 2, min(1, max(0, j + i + traigoUno - 1)));
  }
  memo[{{s, x}, traigoUno}] = res;
  return res;
}
int main() {
  long long int s, x;
  cin >> s >> x;
  cout << dp(s, x) - 2 * (s == x) << endl;
}

```