#include <bits/stdc++.h>
using namespace std;
const int N = 60;
long long n, m, f[N], ans;
void find(int k, long long n, long long sum) {
  if (n < 0 || n > f[k + 2]) return;
  if (!k) {
    if (!n) ans += sum;
    return;
  }
  int x = (m & f[k]) > 0;
  if (x == 1)
    find(k - 1, n - f[k], sum * 2);
  else {
    find(k - 1, n, sum);
    find(k - 1, n - 2 * f[k], sum);
  }
}
int main() {
  cin >> n >> m;
  f[1] = 1;
  for (int i = 2; i < N; i++) f[i] = f[i - 1] * 2;
  find(N - 3, n, 1);
  if (n == m) ans -= 2;
  cout << ans;
}

### Prompt

Please create a solution in CPP 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;
const int N = 60;
long long n, m, f[N], ans;
void find(int k, long long n, long long sum) {
  if (n < 0 || n > f[k + 2]) return;
  if (!k) {
    if (!n) ans += sum;
    return;
  }
  int x = (m & f[k]) > 0;
  if (x == 1)
    find(k - 1, n - f[k], sum * 2);
  else {
    find(k - 1, n, sum);
    find(k - 1, n - 2 * f[k], sum);
  }
}
int main() {
  cin >> n >> m;
  f[1] = 1;
  for (int i = 2; i < N; i++) f[i] = f[i - 1] * 2;
  find(N - 3, n, 1);
  if (n == m) ans -= 2;
  cout << ans;
}

```