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;
const int INF = 1e9 + 7;
int bs(int l, int r, int a, double num) {
  int ans = INF;
  while (l <= r) {
    int mid = (l + r) >> 1;
    double x = num / (mid + 1);
    if (a >= (2 * mid + 1) * x && a <= (2 * mid + 2) * x) {
      ans = mid;
      l = mid + 1;
    } else {
      r = mid - 1;
    }
  }
  return ans;
}
int bs1(int l, int r, int a, double num) {
  int ans = INF;
  while (l <= r) {
    int mid = (l + r) >> 1;
    double x = num / (mid + 1);
    if (a >= (2 * mid) * x && a <= (2 * mid + 1) * x) {
      ans = mid;
      l = mid + 1;
    } else {
      r = mid - 1;
    }
  }
  return ans;
}
int main() {
  int a, b;
  cin >> a >> b;
  double ans1 = (a - b) / 2.0, ans2 = (a + b) / 2.0;
  double x = INF;
  if (ans1 > 0) {
    int maxk = bs1(0, ans1 / b, a, ans1);
    if (maxk != INF) x = min(x, ans1 / maxk);
  }
  int maxk = bs(0, ans2 / b, a, ans2);
  if (maxk != INF) x = min(x, ans2 / (maxk + 1));
  if (x == INF)
    printf("-1\n");
  else
    printf("%.10lf\n", x);
  return 0;
}