#include <bits/stdc++.h>
#pragma GCC optimize("O3")
using namespace std;
long long s, x;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> s >> x;
  if (x > s || (x % 2 != s % 2)) {
    cout << 0;
    return 0;
  }
  long long an = (s - x) / 2;
  long long coef = 1;
  for (int i = 0; i < 50; i++) {
    int bit1 = (an >> i) % 2;
    int bit2 = (x >> i) % 2;
    if (bit1 == 1) {
      if (bit2 != 0) {
        cout << 0;
        return 0;
      }
    } else {
      if (bit2 == 1) {
        coef *= 2;
      }
    }
  }
  if (an == 0) {
    if (x == 0) {
      cout << 0;
      return 0;
    } else {
      coef -= 2;
    }
  }
  cout << coef << endl;
  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>
#pragma GCC optimize("O3")
using namespace std;
long long s, x;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> s >> x;
  if (x > s || (x % 2 != s % 2)) {
    cout << 0;
    return 0;
  }
  long long an = (s - x) / 2;
  long long coef = 1;
  for (int i = 0; i < 50; i++) {
    int bit1 = (an >> i) % 2;
    int bit2 = (x >> i) % 2;
    if (bit1 == 1) {
      if (bit2 != 0) {
        cout << 0;
        return 0;
      }
    } else {
      if (bit2 == 1) {
        coef *= 2;
      }
    }
  }
  if (an == 0) {
    if (x == 0) {
      cout << 0;
      return 0;
    } else {
      coef -= 2;
    }
  }
  cout << coef << endl;
  return 0;
}

```