There is a polyline going through points (0, 0) – (x, x) – (2x, 0) – (3x, x) – (4x, 0) – ... - (2kx, 0) – (2kx + x, x) – .... 

We know that the polyline passes through the point (a, b). Find minimum positive value x such that it is true or determine that there is no such x.

Input

Only one line containing two positive integers a and b (1 ≤ a, b ≤ 109).

Output

Output the only line containing the answer. Your answer will be considered correct if its relative or absolute error doesn't exceed 10 - 9. If there is no such x then output  - 1 as the answer.

Examples

Input

3 1


Output

1.000000000000


Input

1 3


Output

-1


Input

4 1


Output

1.250000000000

Note

You can see following graphs for sample 1 and sample 3. 

<image> <image>
#include <bits/stdc++.h>
using namespace std;
int main() {
  long long x, y;
  cin >> x >> y;
  long long h = x + y;
  for (long long i = (x + y) / (2 * y) + 1; i >= 1; i--) {
    if (h >= 2 * y * i) {
      long long sa = 0, ka = 1, j = 0, l = 0;
      sa = h / (2 * i);
      if (sa != 0) l = 1;
      int u = h % (2 * i);
      while (j < 12) {
        ka *= 10;
        u *= 10;
        ka += (u / (2 * i));
        if (u / (2 * i) != 0) l = 1;
        u %= (2 * i);
        j++;
        if (j >= 9 && l == 0) {
          cout << -1;
          return 0;
        }
      }
      vector<int> v;
      while (ka) {
        v.push_back(ka % 10);
        ka /= 10;
      }
      cout << sa << ".";
      for (int g = v.size() - 2; g >= 0; g--) cout << v[g];
      return 0;
    }
  }
  cout << -1;
}
