#include <bits/stdc++.h>
using namespace std;
const int module = 1e9 + 7;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long s, x;
  cin >> s >> x;
  const int N = 45;
  vector<bool> S(N, 0), X(N, 0);
  for (int i = 0; i < N; ++i) {
    S[i] = s & (1LL << i);
    X[i] = x & (1LL << i);
  }
  vector<long long> dp1(N, 0), dp2(N, 0);
  dp1[0] = 1;
  for (int i = 0; i < N - 1; ++i) {
    if (!S[i] && !X[i]) {
      dp1[i + 1] += dp1[i];
      dp2[i + 1] += dp1[i];
    }
    if (S[i] && X[i]) {
      dp1[i + 1] += 2 * dp1[i];
    }
    if (!S[i] && X[i]) {
      dp2[i + 1] += 2 * dp2[i];
    }
    if (S[i] && !X[i]) {
      dp1[i + 1] += dp2[i];
      dp2[i + 1] += dp2[i];
    }
  }
  cout << dp1[N - 1] - (x == s ? 2 : 0) << "\n";
  return 0;
}

### Prompt

Your challenge is to write 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 int module = 1e9 + 7;
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  long long s, x;
  cin >> s >> x;
  const int N = 45;
  vector<bool> S(N, 0), X(N, 0);
  for (int i = 0; i < N; ++i) {
    S[i] = s & (1LL << i);
    X[i] = x & (1LL << i);
  }
  vector<long long> dp1(N, 0), dp2(N, 0);
  dp1[0] = 1;
  for (int i = 0; i < N - 1; ++i) {
    if (!S[i] && !X[i]) {
      dp1[i + 1] += dp1[i];
      dp2[i + 1] += dp1[i];
    }
    if (S[i] && X[i]) {
      dp1[i + 1] += 2 * dp1[i];
    }
    if (!S[i] && X[i]) {
      dp2[i + 1] += 2 * dp2[i];
    }
    if (S[i] && !X[i]) {
      dp1[i + 1] += dp2[i];
      dp2[i + 1] += dp2[i];
    }
  }
  cout << dp1[N - 1] - (x == s ? 2 : 0) << "\n";
  return 0;
}

```