#include <bits/stdc++.h>
using namespace std;
long long DP[42][2];
bool v[42][2];
long long second, x;
long long solve(int ind, bool carry) {
  if ((1LL << ind) > second) {
    if ((1LL << ind) > x) return (carry == 0);
    return 0;
  }
  long long &temp = DP[ind][carry];
  if (v[ind][carry]) return temp;
  v[ind][carry] = true;
  int a = (((1LL << ind) & x) != 0);
  int b = (((1LL << ind) & second) != 0);
  for (int i = 0; i < 2; i++) {
    for (int j = 0; j < 2; j++) {
      if ((i ^ j) != a) continue;
      if ((i + j + carry) % 2 != b) continue;
      temp += solve(ind + 1, (i + j + carry) / 2);
    }
  }
  return temp;
}
int main() {
  cin >> second >> x;
  cout << solve(0, 0) - (second == x) * 2;
  return 0;
}

### Prompt

Your task is to create 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 DP[42][2];
bool v[42][2];
long long second, x;
long long solve(int ind, bool carry) {
  if ((1LL << ind) > second) {
    if ((1LL << ind) > x) return (carry == 0);
    return 0;
  }
  long long &temp = DP[ind][carry];
  if (v[ind][carry]) return temp;
  v[ind][carry] = true;
  int a = (((1LL << ind) & x) != 0);
  int b = (((1LL << ind) & second) != 0);
  for (int i = 0; i < 2; i++) {
    for (int j = 0; j < 2; j++) {
      if ((i ^ j) != a) continue;
      if ((i + j + carry) % 2 != b) continue;
      temp += solve(ind + 1, (i + j + carry) / 2);
    }
  }
  return temp;
}
int main() {
  cin >> second >> x;
  cout << solve(0, 0) - (second == x) * 2;
  return 0;
}

```