#include <bits/stdc++.h>
using namespace std;
long long S, X;
long long dp[50][50];
long long recur(long long pos, long long co) {
  if (pos == 50LL) return (co == 0LL);
  if (dp[pos][co] != -1LL) return dp[pos][co];
  long long ret = 0LL;
  for (int i = 0; i <= 1; ++i)
    for (int j = 0; j <= 1; ++j) {
      if ((i ^ j) == (X & (1LL << pos)) >> pos &&
          (S & (1LL << pos)) >> pos == ((i + j + co) & 1LL))
        ret += recur(pos + 1LL, (i + j + co) >> 1LL);
    }
  dp[pos][co] = ret;
  return ret;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> S >> X;
  fill_n(dp[0], 50 * 50, -1LL);
  long long ans = recur(0LL, 0LL);
  if ((0LL ^ S) == X) ans -= 2;
  cout << ans << '\n';
  return 0;
}

### Prompt

Create a solution in CPP for 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 S, X;
long long dp[50][50];
long long recur(long long pos, long long co) {
  if (pos == 50LL) return (co == 0LL);
  if (dp[pos][co] != -1LL) return dp[pos][co];
  long long ret = 0LL;
  for (int i = 0; i <= 1; ++i)
    for (int j = 0; j <= 1; ++j) {
      if ((i ^ j) == (X & (1LL << pos)) >> pos &&
          (S & (1LL << pos)) >> pos == ((i + j + co) & 1LL))
        ret += recur(pos + 1LL, (i + j + co) >> 1LL);
    }
  dp[pos][co] = ret;
  return ret;
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cin >> S >> X;
  fill_n(dp[0], 50 * 50, -1LL);
  long long ans = recur(0LL, 0LL);
  if ((0LL ^ S) == X) ans -= 2;
  cout << ans << '\n';
  return 0;
}

```