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() {
  double x, y, i, j, k, a, m = -1;
  cin >> x >> y;
  if (x < 0 || y < 0) {
    cout << -1 << endl;
    return 0;
  }
  cout.precision(10);
  if (y == 0 && x == 0) {
    m = 0;
    cout << fixed << m << endl;
    return 0;
  }
  if (y == 0) {
    m = x / 2;
    cout << fixed << m << endl;
    return 0;
  }
  k = 2.0 * y;
  a = floor((y - x) / k);
  if (x == y) m = y;
  if (a != 0) {
    m = ((y - x) / (2 * a));
    if (m < y) m = -1;
  }
  a = floor((y + x) / (2 * y));
  if (a != 0) {
    if (m != -1) {
      if (m > ((y + x) / (2 * a))) m = ((y + x) / (2 * a));
    } else
      m = ((y + x) / (2 * a));
    if (m < y) m - 1;
  }
  if (m != -1)
    cout << fixed << m << endl;
  else
    cout << -1 << endl;
  return 0;
}