#include <bits/stdc++.h>
using namespace std;
long long DP[50][2];
int main() {
  long long S, X;
  long long ans;
  cin >> S >> X;
  if (S < X) {
    cout << 0 << endl;
    return 0;
  }
  if (S == X)
    ans = -2;
  else
    ans = 0;
  vector<long long> SUM, _XOR;
  while (S) {
    SUM.push_back(S % 2);
    S /= 2;
  }
  for (int i = (0); i < (SUM.size()); i++) {
    _XOR.push_back(X % 2);
    X /= 2;
  }
  SUM.push_back(0);
  _XOR.push_back(0);
  reverse((SUM).begin(), (SUM).end());
  reverse((_XOR).begin(), (_XOR).end());
  memset(DP, 0, sizeof(DP));
  DP[0][0] = 1;
  for (int i = (1); i < (SUM.size() + 1); i++) {
    if (_XOR[i] == 0) {
      if (SUM[i] == 0)
        DP[i][0] += DP[i - 1][0] + DP[i - 1][1];
      else
        DP[i][1] += DP[i - 1][0] + DP[i - 1][1];
    } else {
      if (SUM[i] == 0)
        DP[i][1] += 2LL * DP[i - 1][1];
      else
        DP[i][0] += 2LL * DP[i - 1][0];
    }
  }
  cout << DP[SUM.size() - 1][0] + ans << endl;
}

### 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[50][2];
int main() {
  long long S, X;
  long long ans;
  cin >> S >> X;
  if (S < X) {
    cout << 0 << endl;
    return 0;
  }
  if (S == X)
    ans = -2;
  else
    ans = 0;
  vector<long long> SUM, _XOR;
  while (S) {
    SUM.push_back(S % 2);
    S /= 2;
  }
  for (int i = (0); i < (SUM.size()); i++) {
    _XOR.push_back(X % 2);
    X /= 2;
  }
  SUM.push_back(0);
  _XOR.push_back(0);
  reverse((SUM).begin(), (SUM).end());
  reverse((_XOR).begin(), (_XOR).end());
  memset(DP, 0, sizeof(DP));
  DP[0][0] = 1;
  for (int i = (1); i < (SUM.size() + 1); i++) {
    if (_XOR[i] == 0) {
      if (SUM[i] == 0)
        DP[i][0] += DP[i - 1][0] + DP[i - 1][1];
      else
        DP[i][1] += DP[i - 1][0] + DP[i - 1][1];
    } else {
      if (SUM[i] == 0)
        DP[i][1] += 2LL * DP[i - 1][1];
      else
        DP[i][0] += 2LL * DP[i - 1][0];
    }
  }
  cout << DP[SUM.size() - 1][0] + ans << endl;
}

```