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;
void RI() {}
template <typename... T>
void RI(int& head, T&... tail) {
  scanf("%d", &head);
  RI(tail...);
}
int x, y;
void input() { RI(x, y); }
double solve1() {
  double z = (x - y) / 2.0;
  int r = z / y;
  for (int i = r + 10; i >= r - 10 && i >= 1; i--)
    if (z / i >= y) return z / i;
  return 1001001001;
}
double solve2() {
  double z = (x + y) / 2.0;
  int r = z / y;
  for (int i = r + 10; i >= r - 10 && i >= 1; i--)
    if (z / i >= y) return z / i;
  return 1001001001;
}
void solve() {
  if (x < y) {
    printf("-1\n");
    return;
  }
  printf("%.11f\n", min(solve1(), solve2()));
}
int main() {
  input();
  solve();
  return 0;
}