#include <bits/stdc++.h>
using namespace std;
const int N = 50;
long long din[N][2];
int main() {
  long long a, b;
  cin >> a >> b;
  if ((b & 1) == 1 && (a & 1) == 1) {
    din[0][0] = 2;
  } else if ((b & 1) == 0 & (a & 1) == 0) {
    din[0][0] = 1;
    din[0][1] = 1;
  } else {
    cout << 0 << endl;
    return 0;
  }
  for (int i = 1; i < N; ++i) {
    long long in_xor = (b >> i) & 1;
    long long in_sum = (a >> i) & 1;
    if (in_xor == 1 && in_sum == 0) {
      din[i][1] = din[i - 1][1] * 2;
    } else if (in_xor == 0 && in_sum == 0) {
      din[i][0] = din[i - 1][0];
      din[i][1] = din[i - 1][0];
    } else if (in_xor == 0 && in_sum == 1) {
      din[i][1] = din[i - 1][1];
      din[i][0] = din[i - 1][1];
    } else {
      din[i][0] = din[i - 1][0] * 2;
    }
  }
  long long ans = din[N - 1][0];
  if (a == b) ans -= 2;
  cout << ans << endl;
  return 0;
}

### Prompt

Develop a solution in cpp 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;
const int N = 50;
long long din[N][2];
int main() {
  long long a, b;
  cin >> a >> b;
  if ((b & 1) == 1 && (a & 1) == 1) {
    din[0][0] = 2;
  } else if ((b & 1) == 0 & (a & 1) == 0) {
    din[0][0] = 1;
    din[0][1] = 1;
  } else {
    cout << 0 << endl;
    return 0;
  }
  for (int i = 1; i < N; ++i) {
    long long in_xor = (b >> i) & 1;
    long long in_sum = (a >> i) & 1;
    if (in_xor == 1 && in_sum == 0) {
      din[i][1] = din[i - 1][1] * 2;
    } else if (in_xor == 0 && in_sum == 0) {
      din[i][0] = din[i - 1][0];
      din[i][1] = din[i - 1][0];
    } else if (in_xor == 0 && in_sum == 1) {
      din[i][1] = din[i - 1][1];
      din[i][0] = din[i - 1][1];
    } else {
      din[i][0] = din[i - 1][0] * 2;
    }
  }
  long long ans = din[N - 1][0];
  if (a == b) ans -= 2;
  cout << ans << endl;
  return 0;
}

```