#include <bits/stdc++.h>
using namespace std;
int main() {
  long long a, b;
  cin >> a >> b;
  vector<int> v1, v2;
  while (a) {
    v1.push_back(a % 2);
    a /= 2;
  }
  while (b) {
    v2.push_back(b % 2);
    b /= 2;
  }
  while (v2.size() < v1.size()) v2.push_back(0);
  while (v1.size() < v2.size()) v1.push_back(0);
  int y = 0;
  long long f = 1;
  for (int i = v1.size() - 1; i >= 0; i--) {
    if (i != 0) {
      int u = y + v1[i];
      if (u == 1) {
        if (v2[i] == 1) {
          f *= 2;
          y = 0;
        } else {
          y = 2;
        }
      }
      if (u == 2) {
        if (v2[i] == 0) {
          y = 0;
        } else {
          y = 2;
          f *= 2;
        }
      }
      if (u == 3) {
        if (v2[i] == 1) {
          cout << "0";
          return 0;
        }
        y = 2;
      }
      if (u == 0) {
        if (v2[i] == 1) {
          cout << 0;
          return 0;
        }
      }
    } else {
      int u = y + v1[i];
      if (u == 2 && v2[i] == 1) {
        cout << "0";
        return 0;
      }
      if (u == 3) {
        cout << "0";
        return 0;
      }
      if (u == 1) {
        if (v2[i] == 0) {
          cout << "0";
          return 0;
        }
        f *= 2;
      }
      if (u == 0 && v2[i] == 1) {
        cout << "0";
        return 0;
      }
    }
  }
  int s = 2;
  for (int i = 0; i < v1.size(); i++) {
    if (v1[i] != v2[i]) {
      s = 0;
      break;
    }
  }
  cout << f - s;
}

### Prompt

Construct a CPP code solution to the problem outlined:

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() {
  long long a, b;
  cin >> a >> b;
  vector<int> v1, v2;
  while (a) {
    v1.push_back(a % 2);
    a /= 2;
  }
  while (b) {
    v2.push_back(b % 2);
    b /= 2;
  }
  while (v2.size() < v1.size()) v2.push_back(0);
  while (v1.size() < v2.size()) v1.push_back(0);
  int y = 0;
  long long f = 1;
  for (int i = v1.size() - 1; i >= 0; i--) {
    if (i != 0) {
      int u = y + v1[i];
      if (u == 1) {
        if (v2[i] == 1) {
          f *= 2;
          y = 0;
        } else {
          y = 2;
        }
      }
      if (u == 2) {
        if (v2[i] == 0) {
          y = 0;
        } else {
          y = 2;
          f *= 2;
        }
      }
      if (u == 3) {
        if (v2[i] == 1) {
          cout << "0";
          return 0;
        }
        y = 2;
      }
      if (u == 0) {
        if (v2[i] == 1) {
          cout << 0;
          return 0;
        }
      }
    } else {
      int u = y + v1[i];
      if (u == 2 && v2[i] == 1) {
        cout << "0";
        return 0;
      }
      if (u == 3) {
        cout << "0";
        return 0;
      }
      if (u == 1) {
        if (v2[i] == 0) {
          cout << "0";
          return 0;
        }
        f *= 2;
      }
      if (u == 0 && v2[i] == 1) {
        cout << "0";
        return 0;
      }
    }
  }
  int s = 2;
  for (int i = 0; i < v1.size(); i++) {
    if (v1[i] != v2[i]) {
      s = 0;
      break;
    }
  }
  cout << f - s;
}

```