Problem: 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>

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
const long double eps = 1e-9;
class solver {
 public:
  long double a, b;
  long long ai, bi;
  void input() {
    cin >> ai >> bi;
    a = ai;
    b = bi;
  }
  void solve() {
    if (b > a) {
      cout << -1 << endl;
      return;
    }
    long long nc = ai / bi;
    if (nc % 2 == 0) --nc;
    long double c1 = nc + 1;
    long double c2 = -a;
    long double c3 = b;
    long double xc = (c1 * c3) / (c3 - c2);
    long double yc = (-c1 * c2) / (c3 - c2);
    long double l1 = sqrt(a * a + b * b);
    long double l2 = sqrt(xc * xc + yc * yc);
    long double ans = l1 / l2;
    cout << fixed;
    cout << setprecision(10) << ans << endl;
  }
};
int main() {
  solver slv;
  slv.input();
  slv.solve();
  return 0;
}