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;
long double a, b;
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> a >> b;
  int k1 = (a + b) / (2 * b), k2 = (a - b) / (2 * b);
  long double x1 = -1, x2 = -1;
  if (k1 > 0) {
    x1 = (a + b) / (long double)(k1 * 2);
  }
  if (k2 > 0) {
    x2 = (a - b) / (long double)(k2 * 2);
  }
  cout.precision(16);
  if (x1 <= 0 and x2 <= 0)
    cout << -1 << "\n";
  else if (x1 <= 0) {
    cout << fixed << x2 << "\n";
  } else if (x2 <= 0) {
    cout << fixed << x1 << "\n";
  } else {
    cout << fixed << min(x1, x2) << "\n";
  }
  return 0;
}