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 double EPS = 1e-12;
int solve() {
  double a, b;
  cin >> a >> b;
  double s = a + b;
  double d = a - b;
  double ans1 = (long long)1e18;
  if (d > 0 && (d / (2 * b)) > EPS) {
    double t = floor(d / (2 * b));
    ans1 = d / (2 * t);
  }
  double ans2 = (long long)1e18;
  ;
  if (s / (2 * b) > EPS) {
    double t = floor(s / (2 * b));
    ans2 = s / (2.0 * t);
  }
  double ans = min(ans1, ans2);
  if (abs(ans - (long long)1e18) < EPS) ans = -1;
  cout << setprecision(15) << ans;
  return 0;
}
int main() {
  auto start = chrono::high_resolution_clock::now();
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long test_cases = 1;
  while (test_cases--) solve();
  auto stop = chrono::high_resolution_clock::now();
  auto duration = chrono::duration_cast<chrono::milliseconds>(stop - start);
}