#include <bits/stdc++.h>
using namespace std;
long long dp[50][2];
long long x, s;
long long solve(int p, int c) {
  if (((1LL << p) > s) && ((1LL << p) > x)) return c == 0;
  if (dp[p][c] != -1) return dp[p][c];
  dp[p][c] = 0;
  for (int i = 0; i < (2); i++)
    for (int j = 0; j < (2); j++) {
      int z = i + j + c;
      int sb = (s >> p) & 1LL;
      int xb = (x >> p) & 1LL;
      if (xb == (i ^ j) && sb == (z & 1)) {
        dp[p][c] += solve(p + 1, z / 2);
      }
    }
  return dp[p][c];
}
int main() {
  cin >> s >> x;
  fill(&dp[0][0], &dp[50][0], -1);
  long long ans = solve(0, 0);
  ans -= 2 * (x == s);
  cout << ans << endl;
  return 0;
}

### Prompt

Please create a solution in cpp 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 dp[50][2];
long long x, s;
long long solve(int p, int c) {
  if (((1LL << p) > s) && ((1LL << p) > x)) return c == 0;
  if (dp[p][c] != -1) return dp[p][c];
  dp[p][c] = 0;
  for (int i = 0; i < (2); i++)
    for (int j = 0; j < (2); j++) {
      int z = i + j + c;
      int sb = (s >> p) & 1LL;
      int xb = (x >> p) & 1LL;
      if (xb == (i ^ j) && sb == (z & 1)) {
        dp[p][c] += solve(p + 1, z / 2);
      }
    }
  return dp[p][c];
}
int main() {
  cin >> s >> x;
  fill(&dp[0][0], &dp[50][0], -1);
  long long ans = solve(0, 0);
  ans -= 2 * (x == s);
  cout << ans << endl;
  return 0;
}

```