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 main() {
  ios_base::sync_with_stdio(false);
  srand(time(NULL));
  double a, b;
  cin >> a >> b;
  if (a > b) {
    double res = 0;
    int gauche = 0, droite = a + 1;
    while (gauche < droite) {
      int mid = (gauche + droite + 1) / 2;
      if ((double)(a - b) / (2. * mid) >= b) {
        gauche = mid;
      } else {
        droite = mid - 1;
      }
    }
    res = (double)(a - b) / (2. * gauche);
    gauche = 0;
    droite = a + 1;
    while (gauche < droite) {
      int mid = (gauche + droite + 1) / 2;
      double tmp = (double)(a + b) / (2. * mid);
      if (tmp >= b && (a + b) - 2 * tmp >= 0) {
        gauche = mid;
      } else {
        droite = mid - 1;
      }
    }
    res = min(res, (double)(a + b) / (2. * gauche));
    cout << fixed << setprecision(15) << res;
  } else if (a == b) {
    cout << fixed << setprecision(15) << a;
  } else {
    cout << -1;
  }
}