#include <bits/stdc++.h>
using namespace std;
long long s, x;
int main() {
  cin >> s >> x;
  long long t = s;
  long long cnt = 1;
  for (int i = 0; i < 64; i++) {
    if (x & ((long long)1 << i)) {
      cnt *= 2;
      s -= ((long long)1 << i);
    }
  }
  if (s % 2) {
    cout << 0 << endl;
    return 0;
  }
  s /= 2;
  if (s < 0)
    cout << 0 << endl;
  else {
    for (int i = 0; i < 64; i++) {
      if (s & ((long long)1 << i) && x & ((long long)1 << i)) {
        cout << 0 << endl;
        return 0;
      }
    }
    if (t == x) cnt -= 2;
    cout << cnt << endl;
  }
}

### Prompt

Please provide a CPP coded solution to the problem described below:

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() {
  cin >> s >> x;
  long long t = s;
  long long cnt = 1;
  for (int i = 0; i < 64; i++) {
    if (x & ((long long)1 << i)) {
      cnt *= 2;
      s -= ((long long)1 << i);
    }
  }
  if (s % 2) {
    cout << 0 << endl;
    return 0;
  }
  s /= 2;
  if (s < 0)
    cout << 0 << endl;
  else {
    for (int i = 0; i < 64; i++) {
      if (s & ((long long)1 << i) && x & ((long long)1 << i)) {
        cout << 0 << endl;
        return 0;
      }
    }
    if (t == x) cnt -= 2;
    cout << cnt << endl;
  }
}

```