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;
const int maxn = 100005;
const double eps = 1e-12;
double x, y;
double k;
int cal1(int mid) {
  double temp = (y - x) / (-2 * mid);
  if (temp + eps >= y)
    return 1;
  else
    return 0;
}
int cal2(int mid) {
  double temp = (y + x) / (2 * mid);
  if (temp + eps >= y)
    return 1;
  else
    return 0;
}
int main() {
  while (~scanf("%lf%lf", &x, &y)) {
    k = 1e15;
    int l = 0, r = 1e9 + 5;
    while (l <= r) {
      int mid = (l + r) >> 1;
      if (cal1(mid)) {
        l = mid + 1;
        k = min((y - x) / (-2 * mid), k);
      } else
        r = mid - 1;
    }
    l = 0, r = 1e9 + 5;
    while (l <= r) {
      int mid = (l + r) >> 1;
      if (cal2(mid)) {
        l = mid + 1;
        k = min((y + x) / (2 * mid), k);
      } else
        r = mid - 1;
    }
    if (k > 1e13)
      printf("-1\n");
    else
      printf("%.10lf\n", k);
  }
}
