Problem: 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>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
int main() {
  ios_base::sync_with_stdio(false);
  int a, b;
  cin >> a >> b;
  int k = (a - b) / (2 * b);
  if (a == b) {
    return cout << a, 0;
  }
  double d1 = ((double)a + (double)b) / (2.0 * (double)k + 2.0);
  double d2 = ((double)a - (double)b) / (2.0 * k);
  if (d1 > 0 && d2 > 0) {
    return cout << fixed << setprecision(9) << min(d1, d2) << endl, 0;
  }
  if (k <= 0) {
    return cout << -1, 0;
  }
  if (d1 <= 0 & d2 <= 0) return cout << -1, 0;
  if (d1 > d2) swap(d1, d2);
  if (d1 < 0)
    cout << fixed << setprecision(9) << d2 << endl;
  else
    cout << fixed << setprecision(9) << d1 << endl;
  return 0;
}