#include <bits/stdc++.h>
using namespace std;
inline void prep() {
  cin.tie(0);
  cin.sync_with_stdio(0);
}
const int mod = (int)1e9 + 7;
const int MX = (int)1e6 + 1;
int upd[MX];
int main() {
  prep();
  long long sum, Xor;
  cin >> sum >> Xor;
  if ((sum - Xor) & 1) return puts("0"), 0;
  long long AND = (sum - Xor) >> 1;
  int one_cnt = __builtin_popcountll(Xor);
  if (Xor & AND) return puts("0"), 0;
  long long res = (1ll << one_cnt) - (!AND << 1);
  cout << res << endl;
  return 0;
}

### Prompt

Please provide a Cpp coded solution 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;
inline void prep() {
  cin.tie(0);
  cin.sync_with_stdio(0);
}
const int mod = (int)1e9 + 7;
const int MX = (int)1e6 + 1;
int upd[MX];
int main() {
  prep();
  long long sum, Xor;
  cin >> sum >> Xor;
  if ((sum - Xor) & 1) return puts("0"), 0;
  long long AND = (sum - Xor) >> 1;
  int one_cnt = __builtin_popcountll(Xor);
  if (Xor & AND) return puts("0"), 0;
  long long res = (1ll << one_cnt) - (!AND << 1);
  cout << res << endl;
  return 0;
}

```