#include <bits/stdc++.h>
using namespace std;
long long t, s, x, ans;
int bit[1005];
int main() {
  cin >> s >> x;
  int cnt = 0;
  for (int i = 0; i < 60; i++)
    if (x & (long long)((long long)1 << i)) bit[i] = 1, cnt++;
  t = s - x;
  if (t % 2) {
    puts("0");
    return 0;
  }
  for (int i = 1; i < 60; i++)
    if (t & (long long)((long long)1 << i)) {
      if (bit[i - 1]) {
        puts("0");
        return 0;
      }
    }
  long long ans = 1;
  for (int i = 1; i <= cnt; i++) ans *= 2;
  if (x == s) ans -= 2;
  cout << ans << endl;
}

### 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;
long long t, s, x, ans;
int bit[1005];
int main() {
  cin >> s >> x;
  int cnt = 0;
  for (int i = 0; i < 60; i++)
    if (x & (long long)((long long)1 << i)) bit[i] = 1, cnt++;
  t = s - x;
  if (t % 2) {
    puts("0");
    return 0;
  }
  for (int i = 1; i < 60; i++)
    if (t & (long long)((long long)1 << i)) {
      if (bit[i - 1]) {
        puts("0");
        return 0;
      }
    }
  long long ans = 1;
  for (int i = 1; i <= cnt; i++) ans *= 2;
  if (x == s) ans -= 2;
  cout << ans << endl;
}

```