#include <bits/stdc++.h>
long long int sum, x, dp[105][2];
long long int solve(int dep, int adv) {
  if (dep < 0) return 1 - adv;
  if (dp[dep][adv] != -1) return dp[dep][adv];
  long long int &cur = dp[dep][adv];
  if (x & (1LL << dep)) {
    if (sum & (1LL << dep)) {
      if (adv) return cur = 0LL;
      return cur = solve(dep - 1, 0) * 2LL;
    } else {
      if (adv) return cur = solve(dep - 1, 1) * 2LL;
      return cur = 0LL;
    }
  } else {
    if (sum & (1LL << dep)) return cur = solve(dep - 1, 1);
    return cur = solve(dep - 1, 0);
  }
}
int main() {
  memset(dp, -1, sizeof(dp));
  scanf("%I64d%I64d", &sum, &x);
  printf("%I64d\n", solve(39, 0) - (sum == x ? 2LL : 0LL));
  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>
long long int sum, x, dp[105][2];
long long int solve(int dep, int adv) {
  if (dep < 0) return 1 - adv;
  if (dp[dep][adv] != -1) return dp[dep][adv];
  long long int &cur = dp[dep][adv];
  if (x & (1LL << dep)) {
    if (sum & (1LL << dep)) {
      if (adv) return cur = 0LL;
      return cur = solve(dep - 1, 0) * 2LL;
    } else {
      if (adv) return cur = solve(dep - 1, 1) * 2LL;
      return cur = 0LL;
    }
  } else {
    if (sum & (1LL << dep)) return cur = solve(dep - 1, 1);
    return cur = solve(dep - 1, 0);
  }
}
int main() {
  memset(dp, -1, sizeof(dp));
  scanf("%I64d%I64d", &sum, &x);
  printf("%I64d\n", solve(39, 0) - (sum == x ? 2LL : 0LL));
  return 0;
}

```