#include <bits/stdc++.h>
using namespace std;
const long long LG = 60;
long long dp[LG][2];
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long sum, x;
  cin >> sum >> x;
  dp[0][0] = 1;
  for (long long i = 0; i < LG - 1; ++i) {
    long long want = (sum >> i) & 1;
    long long t = (x >> i) & 1;
    for (long long p = 0; p <= 1; ++p) {
      for (long long pr = 0; pr <= 1; ++pr) {
        for (long long a = 0; a <= 1; ++a) {
          for (long long b = 0; b <= 1; ++b) {
            if ((a + b + pr) % 2 == want && (a ^ b) == t &&
                (a + b + pr) / 2 == p) {
              dp[i + 1][p] += dp[i][pr];
            }
          }
        }
      }
    }
  }
  cout << dp[LG - 1][0] - ((sum == x) << 1) << '\n';
  return 0;
}

### Prompt

Your task is to create 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;
const long long LG = 60;
long long dp[LG][2];
signed main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long sum, x;
  cin >> sum >> x;
  dp[0][0] = 1;
  for (long long i = 0; i < LG - 1; ++i) {
    long long want = (sum >> i) & 1;
    long long t = (x >> i) & 1;
    for (long long p = 0; p <= 1; ++p) {
      for (long long pr = 0; pr <= 1; ++pr) {
        for (long long a = 0; a <= 1; ++a) {
          for (long long b = 0; b <= 1; ++b) {
            if ((a + b + pr) % 2 == want && (a ^ b) == t &&
                (a + b + pr) / 2 == p) {
              dp[i + 1][p] += dp[i][pr];
            }
          }
        }
      }
    }
  }
  cout << dp[LG - 1][0] - ((sum == x) << 1) << '\n';
  return 0;
}

```