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;
double Min(double x, double y) { return x < y ? x : y; }
int main() {
  double a, b;
  double aab, asb, k1, k2;
  while (cin >> a >> b) {
    aab = a + b;
    asb = a - b;
    if (asb == 0) {
      printf("%.12lf\n", b);
      continue;
    }
    if (asb < 0) {
      printf("-1\n");
      continue;
    }
    k1 = floor(aab / 2.0 / b);
    k2 = floor(asb / 2.0 / b);
    printf("%.12lf\n", Min(aab / 2.0 / k1, asb / 2.0 / k2));
  }
  return 0;
}
