#include <bits/stdc++.h>
using namespace std;
long long int x, s;
long long int dp[70][2];
long long int solve(int pos, int curr) {
  if (pos >= 61) return curr == ((s >> pos) & 1);
  if (dp[pos][curr] != -1) return dp[pos][curr];
  long long int ret = 0;
  if ((x >> pos) & 1) {
    if ((s >> pos) & 1) {
      if (curr == 1)
        return 0;
      else
        ret = solve(pos + 1, 0) * 2;
    } else {
      if (curr == 0)
        return 0;
      else
        ret = solve(pos + 1, 1) * 2;
    }
  } else {
    if ((s >> pos) & 1) {
      if (curr == 1)
        ret = solve(pos + 1, 0) + solve(pos + 1, 1);
      else
        return 0;
    } else {
      if (curr == 0)
        ret = solve(pos + 1, 1) + solve(pos + 1, 0);
      else
        return 0;
    }
  }
  dp[pos][curr] = ret;
  return ret;
}
int main() {
  memset((dp), (-1), sizeof(dp));
  cin >> s >> x;
  cout << solve(0, 0) - 2 * (s == x) << "\n";
  return 0;
}

### Prompt

Please formulate 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;
long long int x, s;
long long int dp[70][2];
long long int solve(int pos, int curr) {
  if (pos >= 61) return curr == ((s >> pos) & 1);
  if (dp[pos][curr] != -1) return dp[pos][curr];
  long long int ret = 0;
  if ((x >> pos) & 1) {
    if ((s >> pos) & 1) {
      if (curr == 1)
        return 0;
      else
        ret = solve(pos + 1, 0) * 2;
    } else {
      if (curr == 0)
        return 0;
      else
        ret = solve(pos + 1, 1) * 2;
    }
  } else {
    if ((s >> pos) & 1) {
      if (curr == 1)
        ret = solve(pos + 1, 0) + solve(pos + 1, 1);
      else
        return 0;
    } else {
      if (curr == 0)
        ret = solve(pos + 1, 1) + solve(pos + 1, 0);
      else
        return 0;
    }
  }
  dp[pos][curr] = ret;
  return ret;
}
int main() {
  memset((dp), (-1), sizeof(dp));
  cin >> s >> x;
  cout << solve(0, 0) - 2 * (s == x) << "\n";
  return 0;
}

```