#include <bits/stdc++.h>
using namespace std;
long long s, x;
int main() {
  ios_base::sync_with_stdio(0);
  cin >> s >> x;
  if ((s - x) & 1) return cout << 0, 0;
  long long left = (s - x) / 2;
  for (int i = 60; i >= 0; i--) {
    long long pwr = 1LL << i;
    if (left >= pwr) {
      left -= pwr;
      if (x & (1LL << i)) return cout << 0, 0;
    }
  }
  if (left != 0) return cout << 0, 0;
  int cnt = 0;
  long long n = x;
  while (n) {
    if (n & 1) cnt++;
    n >>= 1;
  }
  long long res = 1LL << cnt;
  if (s == x) res -= 2;
  cout << res;
  return 0;
}

### Prompt

Please formulate a CPP solution to 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 s, x;
int main() {
  ios_base::sync_with_stdio(0);
  cin >> s >> x;
  if ((s - x) & 1) return cout << 0, 0;
  long long left = (s - x) / 2;
  for (int i = 60; i >= 0; i--) {
    long long pwr = 1LL << i;
    if (left >= pwr) {
      left -= pwr;
      if (x & (1LL << i)) return cout << 0, 0;
    }
  }
  if (left != 0) return cout << 0, 0;
  int cnt = 0;
  long long n = x;
  while (n) {
    if (n & 1) cnt++;
    n >>= 1;
  }
  long long res = 1LL << cnt;
  if (s == x) res -= 2;
  cout << res;
  return 0;
}

```