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;
using ld = long double;
int main() {
  ios::sync_with_stdio(false), cin.tie(0);
  cout << fixed << setprecision(10);
  int a, b;
  cin >> a >> b;
  if (a == b) {
    cout << a << endl;
    return 0;
  }
  if (a < b) {
    cout << -1 << endl;
    return 0;
  }
  ld res = a + b;
  if (a - b >= b * 2) {
    int c = a - b;
    int lb = 1, ub = 1e9 + 1000;
    while (ub - lb > 1) {
      int m = (lb + ub) / 2;
      if (c < 1LL * m * 2 * b) {
        ub = m;
      } else {
        lb = m;
      }
    }
    res = min(res, (ld)c / (lb * 2));
  }
  {
    int c = a + b;
    int lb = 1, ub = 1e9 + 1000;
    while (ub - lb > 1) {
      int m = (lb + ub) / 2;
      if (c < 1LL * m * 2 * b) {
        ub = m;
      } else {
        lb = m;
      }
    }
    res = min(res, (ld)c / (lb * 2));
  }
  cout << res << endl;
  return 0;
}