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;
double a, b;
int x, cnt;
double dmin(double a, double b) { return (a < b) ? a : b; }
int main() {
  cin >> a >> b;
  if (a < b) {
    cout << -1 << endl;
    return 0;
  }
  if (a == b) {
    cout << a << endl;
    return 0;
  }
  int l = 1;
  int r = 2e9;
  double ans;
  while (l <= r) {
    int mid = (l + r) / 2;
    double x = (a + b) / (2.0 * mid);
    if (x >= b) {
      l = mid + 1;
    } else {
      r = mid - 1;
    }
  }
  if ((a + b) / (2.0 * (l - 1)) >= b)
    ans = (a + b) / (2.0 * (l - 1));
  else
    ans = (a + b) / (2.0 * l);
  l = 1;
  r = 2e9;
  while (l <= r) {
    int mid = (l + r) / 2;
    double x = (a - b) / (2.0 * mid);
    if (x >= b) {
      l = mid + 1;
    } else {
      r = mid - 1;
    }
  }
  if ((a + b) / (2.0 * (l - 1)) >= b)
    ans = dmin((a + b) / (2.0 * (l - 1)), ans);
  else
    ans = dmin((a - b) / (2.0 * l), ans);
  cout << setiosflags(ios::fixed) << setprecision(9) << ans << endl;
  return 0;
}