#include <bits/stdc++.h>
using namespace std;
long long dp[55][2];
int main() {
  long long X, Y;
  scanf("%lld %lld", &Y, &X);
  dp[51][0] = 1;
  for (int i = 50; i >= 0; i--) {
    for (int flag = 0; flag <= 1; flag++) {
      bool x = (X >> i) & 1, y = (Y >> i) & 1;
      if (x)
        dp[i][flag] += dp[i + 1][(flag + 1) / 2] * (flag != y) * 2;
      else {
        dp[i][flag] += dp[i + 1][(flag + 2) / 2] * (flag == y);
        dp[i][flag] += dp[i + 1][0] * (flag == y);
      }
    }
  }
  if (!X) assert(dp[0][0] - (X == Y) * 2 == (!(Y & 1)));
  printf("%lld\n", dp[0][0] - (X == Y) * 2);
  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 dp[55][2];
int main() {
  long long X, Y;
  scanf("%lld %lld", &Y, &X);
  dp[51][0] = 1;
  for (int i = 50; i >= 0; i--) {
    for (int flag = 0; flag <= 1; flag++) {
      bool x = (X >> i) & 1, y = (Y >> i) & 1;
      if (x)
        dp[i][flag] += dp[i + 1][(flag + 1) / 2] * (flag != y) * 2;
      else {
        dp[i][flag] += dp[i + 1][(flag + 2) / 2] * (flag == y);
        dp[i][flag] += dp[i + 1][0] * (flag == y);
      }
    }
  }
  if (!X) assert(dp[0][0] - (X == Y) * 2 == (!(Y & 1)));
  printf("%lld\n", dp[0][0] - (X == Y) * 2);
  return 0;
}

```