#include <bits/stdc++.h>
using namespace std;
int main() {
  long long s, x;
  cin >> s >> x;
  s -= x;
  vector<long long> nums;
  long long ans = 1;
  for (long long i = 1, j = 0; j < 41; i *= 2, j++) {
    if (x & 1)
      ans *= 2;
    else
      nums.push_back(2 * i);
    x >>= 1;
  }
  reverse(nums.begin(), nums.end());
  bool took = false;
  for (long long i : nums) {
    if (s >= i) {
      took = true;
      s -= i;
    }
  }
  if (!took) ans -= 2;
  if (s == 0)
    cout << ans << '\n';
  else
    cout << 0 << '\n';
}

### Prompt

Construct a CPP code solution to the problem outlined:

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;
int main() {
  long long s, x;
  cin >> s >> x;
  s -= x;
  vector<long long> nums;
  long long ans = 1;
  for (long long i = 1, j = 0; j < 41; i *= 2, j++) {
    if (x & 1)
      ans *= 2;
    else
      nums.push_back(2 * i);
    x >>= 1;
  }
  reverse(nums.begin(), nums.end());
  bool took = false;
  for (long long i : nums) {
    if (s >= i) {
      took = true;
      s -= i;
    }
  }
  if (!took) ans -= 2;
  if (s == 0)
    cout << ans << '\n';
  else
    cout << 0 << '\n';
}

```