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 = 0.00000000000001;
long a;
long b;
long double f(long double x, long n) {
  long double x1, x2, y1, y2;
  y1 = x;
  y2 = 0;
  x1 = (2 * n - 1) * x;
  x2 = 2 * n * x;
  long double d1S = (a - x1) * (a - x1) + (b - y1) * (b - y1);
  long double d2S = (a - x2) * (a - x2) + (b - y2) * (b - y2);
  long double yS =
      (2 * x * x + d1S - d2S) * (2 * x * x + d1S - d2S) / 8.0 / x / x;
  long double res = sqrt(d1S - yS);
  return res;
}
int main() {
  cout << fixed;
  cout << setprecision(10);
  cin >> a >> b;
  if (b > a)
    cout << -1 << endl;
  else if (a == b)
    cout << a << endl;
  else {
    long n = ((a - b) / (2 * b)) + 1;
    if (n == 1)
      cout << (a + b) / 2.0 << endl;
    else {
      long double l = b;
      long double h = a / 2.0 / (n - 1);
      long double m1 = l;
      long double m2 = h;
      long double res1 = 1;
      long double res2 = 0;
      for (int i = 0; i <= 10000; i++) {
        m1 = (2 * l + h) / 3;
        m2 = (2 * h + l) / 3;
        res1 = f(m1, n);
        res2 = f(m2, n);
        if (res1 < res2)
          h = m2;
        else
          l = m1;
      }
      cout << (m1 + m2) / 2 << endl;
    }
  }
}