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 a, b;
double x;
int main(void) {
  scanf("%lf%lf", &a, &b);
  if (b > a) {
    printf("-1\n");
    return 0;
  }
  int k = 1;
  while (2 * k * b + b < a) {
    k++;
  }
  int point = 2 * k * b + b;
  if (point == a) {
    printf("%.12lf\n", b);
    return 0;
  }
  k--;
  point = 2 * k * b + b;
  double incl = b / point;
  double numerador = a + b;
  double denominador = 1.0 / incl + 1;
  x = numerador / denominador;
  printf("%.12lf\n", x);
  return 0;
}
