#include <bits/stdc++.h>
using namespace std;
void solve() {
  long long n, m;
  cin >> n >> m;
  if ((n - m) % 2) {
    cout << 0 << '\n';
    return;
  }
  long long dif = (n - m) / 2;
  long long ans = 1;
  for (long long i = 50; i >= 0; i--) {
    if (m & (1LL << i)) ans *= 2;
    if (dif & (1ll << i)) {
      if (m & (1ll << i)) {
        cout << 0 << '\n';
        return;
      }
    }
  }
  if (n == m) ans -= 2;
  cout << ans << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t = 1;
  while (t--) {
    solve();
  }
}

### Prompt

Generate 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;
void solve() {
  long long n, m;
  cin >> n >> m;
  if ((n - m) % 2) {
    cout << 0 << '\n';
    return;
  }
  long long dif = (n - m) / 2;
  long long ans = 1;
  for (long long i = 50; i >= 0; i--) {
    if (m & (1LL << i)) ans *= 2;
    if (dif & (1ll << i)) {
      if (m & (1ll << i)) {
        cout << 0 << '\n';
        return;
      }
    }
  }
  if (n == m) ans -= 2;
  cout << ans << '\n';
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t = 1;
  while (t--) {
    solve();
  }
}

```