#include <bits/stdc++.h>
using namespace std;
int arr1[2100000];
int arr2[2100000];
int main() {
  long long s, x, c = 0;
  cin >> s >> x;
  long long t = s - x;
  for (int i = 0; i <= 60; i++) {
    if ((1LL << (60 - i)) & x) {
      c++;
    } else {
      if (t >= (1LL << (61 - i))) {
        t -= (1LL << (61 - i));
      }
    }
  }
  if (t != 0)
    cout << 0 << endl;
  else if (s == x) {
    cout << (1LL << (c)) - 2 << endl;
  } else {
    cout << (1LL << (c)) << endl;
  }
  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;
int arr1[2100000];
int arr2[2100000];
int main() {
  long long s, x, c = 0;
  cin >> s >> x;
  long long t = s - x;
  for (int i = 0; i <= 60; i++) {
    if ((1LL << (60 - i)) & x) {
      c++;
    } else {
      if (t >= (1LL << (61 - i))) {
        t -= (1LL << (61 - i));
      }
    }
  }
  if (t != 0)
    cout << 0 << endl;
  else if (s == x) {
    cout << (1LL << (c)) - 2 << endl;
  } else {
    cout << (1LL << (c)) << endl;
  }
  return 0;
}

```