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 e = 0.00000000000000001;
int main() {
  ios_base::sync_with_stdio(0);
  int x, y;
  cin >> x >> y;
  if (x < y) {
    cout << -1 << endl;
    return 0;
  }
  if (x == y) {
    cout << x << ".000000000" << endl;
    return 0;
  }
  if (x % y == 0 && (x / y) % 2 == 1) {
    cout << y << ".000000000" << endl;
    return 0;
  }
  double X = ((double)x + y);
  double ans = 2000000000.0;
  double X1 = ((double(x) - y));
  if (X1 > 0) {
    int l = 1, r = (int)X1;
    while (l <= r) {
      int q = (l + r) / 2;
      double R = X1 / q;
      R /= 2;
      if (R >= (double)y) {
        ans = min(ans, R);
        l = q + 1;
      } else
        r = q - 1;
    }
  }
  int l = 1, r = (int)X;
  while (l <= r) {
    int q = (l + r) / 2;
    double R = X / q;
    R /= 2;
    if (R >= (double)y) {
      ans = min(ans, R);
      l = q + 1;
    } else
      r = q - 1;
  }
  cout << fixed << setprecision(9) << ans << endl;
  return 0;
}