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 a, b, x, f = 0;
  double ans1 = 1000000001, ans2 = 1000000001, t;
  scanf("%lld %lld", &a, &b);
  if (a < b) {
    cout << -1;
    return 0;
  }
  if (a == b) {
    cout << b;
    return 0;
  }
  x = a - b;
  if (x >= 2 * b) {
    x = (a - b) / b;
    if (x % 2 == 0) {
      t = x;
      t = (a - b) / t;
      ans1 = t;
    } else {
      x--;
      t = x;
      t = (a - b) / t;
      ans1 = t;
    }
  }
  x = a + b;
  if (x >= 2 * b) {
    x = (a + b) / b;
    if (x % 2 == 0) {
      t = x;
      t = (a + b) / t;
      ans2 = t;
    } else {
      x--;
      t = x;
      t = (a + b) / t;
      ans2 = t;
    }
  }
  if (min(ans1, ans2) == 1000000001) {
    cout << -1;
    return 0;
  }
  cout << fixed << std::setprecision(15) << min(ans1, ans2);
  return 0;
}
