#include <bits/stdc++.h>
using namespace std;
long long sum, Xor;
long long solve(int p, int carry, long long m) {
  if (p > m) return !carry;
  int s = (sum >> p) & 1;
  int x = (Xor >> p++) & 1;
  if (!s) {
    if (!x && !carry)
      return solve(p, 0, m) + solve(p, 1, m);
    else if (x && carry)
      return 2 * solve(p, 1, m);
  } else {
    if (!x && carry)
      return solve(p, 0, m) + solve(p, 1, m);
    else if (x && !carry)
      return 2 * solve(p, 0, m);
  }
  return 0;
}
int main() {
  cin >> sum >> Xor;
  if (sum < Xor)
    cout << 0;
  else
    cout << solve(0, 0, log2(sum)) - 2 * (sum == Xor);
  return 0;
}

### 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 sum, Xor;
long long solve(int p, int carry, long long m) {
  if (p > m) return !carry;
  int s = (sum >> p) & 1;
  int x = (Xor >> p++) & 1;
  if (!s) {
    if (!x && !carry)
      return solve(p, 0, m) + solve(p, 1, m);
    else if (x && carry)
      return 2 * solve(p, 1, m);
  } else {
    if (!x && carry)
      return solve(p, 0, m) + solve(p, 1, m);
    else if (x && !carry)
      return 2 * solve(p, 0, m);
  }
  return 0;
}
int main() {
  cin >> sum >> Xor;
  if (sum < Xor)
    cout << 0;
  else
    cout << solve(0, 0, log2(sum)) - 2 * (sum == Xor);
  return 0;
}

```