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;
long long a, b, r, e, k;
double res1, res2, res;
bool co1, co2;
int main() {
  ios::sync_with_stdio(0);
  scanf("%d %d", &a, &b);
  if (a == b) {
    double p = double(a);
    printf("%0.9lf", p);
    return 0;
  }
  res1 = res2 = 5000000000;
  r = a - b;
  e = a + b;
  co1 = co2 = true;
  if (e / b >= 2) {
    k = e / b;
    if (k % 2) k--;
    res1 = e / double(k);
  } else
    co1 = false;
  if (r > 0) {
    e = r;
    if (e / b >= 2) {
      k = e / b;
      if (k % 2) k--;
      res2 = e / double(k);
    } else
      co2 = false;
  } else
    co2 = false;
  if (!co1 && !co2)
    printf("-1");
  else {
    res = min(res1, res2);
    printf("%0.9f", res);
  }
  return 0;
}