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 double exps = 1e-10;
const int INF = 0x3f3f3f3f;
int to_int(double x) {
  int l = x, r = x + 1;
  if (r - x < exps) return r;
  return l;
}
int main() {
  double a, b;
  while (~scanf("%lf%lf", &a, &b)) {
    if (a < b) {
      printf("-1\n");
      continue;
    }
    double ans = INF;
    int k;
    k = to_int((a - b) / (2 * b));
    if (k > 0) ans = (a - b) / (2.0 * k);
    ans = min(ans, (a + b) / (2.0 * (k + 1)));
    printf("%.12lf\n", ans);
  }
  return 0;
}
