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() {
  long long a, b, t1, t2;
  double ans1, ans2, ans;
  scanf("%I64d%I64d", &a, &b);
  if (a < b) {
    printf("-1\n");
    return 0;
  } else if (a == b) {
    printf("%d\n", b);
    return 0;
  }
  t1 = (((a + b) / b) & 1 ? (a + b) / b - 1 : (a + b) / b);
  ans = ans1 = (double)(a + b) / (double)t1;
  if (a - b > b) {
    t2 = (((a - b) / b) & 1 ? (a - b) / b - 1 : (a - b) / b);
    ans2 = (double)(a - b) / (double)t2;
    ans = min(ans1, ans2);
  }
  printf("%.12f\n", ans);
  return 0;
}
