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() {
  int x, y;
  scanf("%d %d", &x, &y);
  if (y > x) {
    printf("-1\n");
    return 0;
  }
  int left = 1, right = 1000000000;
  double top = (x + y) / 2.0;
  while (left + 1 < right) {
    int mid = (left + right) / 2;
    if (top / mid >= y) {
      left = mid;
    } else {
      right = mid;
    }
  }
  double ans = top / left;
  if ((x - y) / 2.0 >= y) {
    top = (x - y) / 2.0;
    left = 1, right = 1000000000;
    while (left + 1 < right) {
      int mid = (left + right) / 2;
      if (top / mid >= y) {
        left = mid;
      } else {
        right = mid;
      }
    }
    ans = min(ans, top / left);
  }
  printf("%.10lf\n", ans);
  return 0;
}
