#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  long long sum, xr;
  cin >> sum >> xr;
  long long dp[2] = {1, 0};
  for (int bit = 0; bit < 50; bit++) {
    long long ndp[2] = {0, 0};
    int needSum = (sum >> bit) & 1;
    int needXor = (xr >> bit) & 1;
    for (int was = 0; was < 2; was++) {
      for (int b1 = 0; b1 < 2; b1++) {
        for (int b2 = 0; b2 < 2; b2++) {
          int mySum = (b1 + b2 + was) % 2;
          int myXor = b1 ^ b2;
          if (myXor == needXor && mySum == needSum) {
            ndp[(b1 + b2 + was) >> 1] += dp[was];
          }
        }
      }
    }
    for (int _n(2), i(0); i < _n; i++) dp[i] = ndp[i];
  }
  cout << (dp[0] - (sum == xr ? 2 : 0)) << endl;
  return 0;
}

### Prompt

In Cpp, your task is to solve 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;
int main() {
  ios_base::sync_with_stdio(false);
  long long sum, xr;
  cin >> sum >> xr;
  long long dp[2] = {1, 0};
  for (int bit = 0; bit < 50; bit++) {
    long long ndp[2] = {0, 0};
    int needSum = (sum >> bit) & 1;
    int needXor = (xr >> bit) & 1;
    for (int was = 0; was < 2; was++) {
      for (int b1 = 0; b1 < 2; b1++) {
        for (int b2 = 0; b2 < 2; b2++) {
          int mySum = (b1 + b2 + was) % 2;
          int myXor = b1 ^ b2;
          if (myXor == needXor && mySum == needSum) {
            ndp[(b1 + b2 + was) >> 1] += dp[was];
          }
        }
      }
    }
    for (int _n(2), i(0); i < _n; i++) dp[i] = ndp[i];
  }
  cout << (dp[0] - (sum == xr ? 2 : 0)) << endl;
  return 0;
}

```