#include <bits/stdc++.h>
using namespace std;
long long S1, S, X;
bool has[60];
int main() {
  ios::sync_with_stdio(0);
  cin >> S1 >> X;
  S = S1;
  long long ans = 1;
  for (int i = 0; i < 60; i++) {
    long long b = ((long long)1) << i;
    if (X & b) {
      S -= b;
      ans <<= 1;
    } else
      has[i] = true;
  }
  if (S < 0 || S % 2 == 1) {
    cout << "0\n";
    return 0;
  }
  S /= 2;
  for (int i = 0; i < 60; i++) {
    long long b = ((long long)1) << i;
    if ((S & b) && !has[i]) {
      cout << "0\n";
      return 0;
    }
  }
  if (S1 == X) ans -= 2;
  cout << ans << "\n";
  return 0;
}

### Prompt

In Cpp, your task is to solve 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 S1, S, X;
bool has[60];
int main() {
  ios::sync_with_stdio(0);
  cin >> S1 >> X;
  S = S1;
  long long ans = 1;
  for (int i = 0; i < 60; i++) {
    long long b = ((long long)1) << i;
    if (X & b) {
      S -= b;
      ans <<= 1;
    } else
      has[i] = true;
  }
  if (S < 0 || S % 2 == 1) {
    cout << "0\n";
    return 0;
  }
  S /= 2;
  for (int i = 0; i < 60; i++) {
    long long b = ((long long)1) << i;
    if ((S & b) && !has[i]) {
      cout << "0\n";
      return 0;
    }
  }
  if (S1 == X) ans -= 2;
  cout << ans << "\n";
  return 0;
}

```