#include <bits/stdc++.h>
using namespace std;
long long Rec(long long bit, long long y);
long long s, x;
long long Dp[60][2];
int main() {
  memset(Dp, -1, sizeof(Dp));
  cin >> s >> x;
  cout << Rec(0, 0) - 2LL * (s == x);
}
long long Rec(long long bit, long long y) {
  if (bit == 45) return (y == 0);
  if (Dp[bit][y] != -1) return Dp[bit][y];
  long long Res = 0;
  long long Sb = ((s & (1LL << bit)) != 0LL);
  long long Xb = ((x & (1LL << bit)) != 0LL);
  if (!Xb && y == Sb) Res += Rec(bit + 1, 0) + Rec(bit + 1, 1);
  if (Xb && ((y + 1) & 1) == Sb) Res += 2 * Rec(bit + 1, y > 0 ? 1 : 0);
  return Dp[bit][y] = Res;
}

### 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;
long long Rec(long long bit, long long y);
long long s, x;
long long Dp[60][2];
int main() {
  memset(Dp, -1, sizeof(Dp));
  cin >> s >> x;
  cout << Rec(0, 0) - 2LL * (s == x);
}
long long Rec(long long bit, long long y) {
  if (bit == 45) return (y == 0);
  if (Dp[bit][y] != -1) return Dp[bit][y];
  long long Res = 0;
  long long Sb = ((s & (1LL << bit)) != 0LL);
  long long Xb = ((x & (1LL << bit)) != 0LL);
  if (!Xb && y == Sb) Res += Rec(bit + 1, 0) + Rec(bit + 1, 1);
  if (Xb && ((y + 1) & 1) == Sb) Res += 2 * Rec(bit + 1, y > 0 ? 1 : 0);
  return Dp[bit][y] = Res;
}

```