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;
int a, b;
double Xcu, Xms;
double len() {
  if (Xcu / 2 < b) return -1;
  int l = 0, r = 1000000007;
  while (r - l > 1) {
    int middle = ((l + r) >> 1);
    double x = Xcu / (middle * 2);
    if (x >= b)
      l = middle;
    else
      r = middle;
  }
  return Xcu / (l * 2);
}
double xuong() {
  if (Xms / 2 < b) return -1;
  int l = 0, r = 1000000007;
  while (r - l > 1) {
    int middle = ((l + r) >> 1);
    double x = Xms / (middle * 2);
    if (x >= b)
      l = middle;
    else
      r = middle;
  }
  return Xms / (l * 2);
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> a >> b;
  Xcu = a - b, Xms = a + b;
  double x = len();
  double y = xuong();
  if (x == -1 && y == -1) {
    cout << -1;
    return 0;
  }
  if (x == -1) {
    cout << fixed << setprecision(9) << y;
    return 0;
  }
  if (y == -1) {
    cout << fixed << setprecision(9) << x;
    return 0;
  }
  cout << fixed << setprecision(9) << min(x, y);
  return 0;
}