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>
#include <bits/stdc++.h>
using namespace std;
double a, b;
bool ck(double x) {
  double ta = a / x;
  int t = round(ta);
  double k = x * t;
  k = (a - k);
  if (t & 1) {
    if (fabs((k - (x - b))) <= 1e-9)
      return true;
    else
      return false;
  }
  if (fabs((k - (b))) <= 1e-9)
    return true;
  else
    return false;
}
int main() {
  int i, j, k;
  double l, r, mid;
  cin >> a >> b;
  if (b > a) {
    cout << -1 << endl;
    return 0;
  }
  if (ck(b) == true) {
    printf("%.9lf\n", b);
    return 0;
  }
  int nl = 0, nr = 1000000000;
  double x1;
  while (nl != nr) {
    int midl = (nl + nr + 1) / 2;
    x1 = (double)(a + b) / (double)(2 * midl + 2);
    if (x1 >= b) {
      nl = midl;
    } else {
      nr = midl - 1;
    }
  }
  double mini = (double)(a + b) / (double)(2 * nl + 2);
  printf("%.9lf\n", mini);
  return 0;
}
