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>
double a, b;
double fun(double x) {
  if (x / 2 < b) return 1000000001;
  int ctr1 = 1;
  int ctr2 = 1000000000;
  while (1) {
    if (ctr1 == ctr2) break;
    int num1 = (ctr1 + ctr2 + 1) / 2;
    if (x / (2 * num1) < b) {
      ctr2 = num1 - 1;
    } else {
      ctr1 = num1;
    }
  }
  return x / (2 * ctr1);
}
int main() {
  scanf("%lf%lf", &a, &b);
  if (a < b) {
    printf("-1");
    return 0;
  }
  if (a == b) {
    printf("%.10lf\n", double(a));
    return 0;
  }
  double x = double(a - b);
  double y = double(a + b);
  if (x >= 2)
    x = fun(x);
  else
    x = 1000000001;
  y = fun(y);
  if (x < y)
    printf("%.10lf", x);
  else
    printf("%.10lf", y);
  return 0;
}
