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() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int a, b, c, temp1, temp2;
  double d, x1 = -1, x2 = -1;
  cin >> a >> b;
  temp1 = (a - b) / (2 * b);
  temp2 = (a + b) / (2 * b);
  if (temp1 > 0) x1 = (a - b) / (2.0 * temp1);
  if (temp2 > 0) x2 = (a + b) / (2.0 * temp2);
  if (x1 <= 0 and x2 <= 0)
    cout << "-1\n";
  else if (x1 <= 0)
    printf("%0.9lf\n", x2);
  else if (x2 <= 0)
    printf("%0.9lf\n", x1);
  else
    printf("%0.9lf\n", min(x1, x2));
  return 0;
}
