#include <bits/stdc++.h>
using namespace std;
const int maxN = 200009;
vector<int> a;
vector<int> b;
vector<int> bb;
int main() {
  long long s, x;
  cin >> s >> x;
  int s1 = s, x1 = x;
  int ans = 0;
  for (int i = 0; i < 45; i++) {
    if ((s & ((long long)1 << i)) && (x & ((long long)1 << i))) ans++;
    if (!(s & ((long long)1 << i)) && (x & ((long long)1 << i))) {
      cout << 0;
      return 0;
    }
    if ((s & ((long long)1 << i)) && !(x & ((long long)1 << i))) {
      cout << 0;
      return 0;
    }
    if (!(s & ((long long)1 << i)) && !(x & ((long long)1 << i))) {
      if ((s & ((long long)1 << (i + 1))) && !(x & ((long long)1 << (i + 1)))) {
        s -= ((long long)1 << (i + 1));
      }
      if (!(s & ((long long)1 << (i + 1))) && (x & ((long long)1 << (i + 1)))) {
        s -= ((long long)1 << (i + 1));
      }
    }
  }
  if (s1 != x1)
    cout << ((long long)1 << ans);
  else
    cout << ((long long)1 << ans) - 2;
  return 0;
}

### 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;
const int maxN = 200009;
vector<int> a;
vector<int> b;
vector<int> bb;
int main() {
  long long s, x;
  cin >> s >> x;
  int s1 = s, x1 = x;
  int ans = 0;
  for (int i = 0; i < 45; i++) {
    if ((s & ((long long)1 << i)) && (x & ((long long)1 << i))) ans++;
    if (!(s & ((long long)1 << i)) && (x & ((long long)1 << i))) {
      cout << 0;
      return 0;
    }
    if ((s & ((long long)1 << i)) && !(x & ((long long)1 << i))) {
      cout << 0;
      return 0;
    }
    if (!(s & ((long long)1 << i)) && !(x & ((long long)1 << i))) {
      if ((s & ((long long)1 << (i + 1))) && !(x & ((long long)1 << (i + 1)))) {
        s -= ((long long)1 << (i + 1));
      }
      if (!(s & ((long long)1 << (i + 1))) && (x & ((long long)1 << (i + 1)))) {
        s -= ((long long)1 << (i + 1));
      }
    }
  }
  if (s1 != x1)
    cout << ((long long)1 << ans);
  else
    cout << ((long long)1 << ans) - 2;
  return 0;
}

```