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>
#include <bits/stdc++.h>
using namespace std;
long long a, b;
double solve(long long len) {
  long long l = 1, r = 1000000000LL;
  while (r - l > 1) {
    long long mid = (l + r) / 2;
    if (len >= b * 2 * mid)
      l = mid;
    else
      r = mid - 1;
  }
  while (len < b * 2 * r) r--;
  return (double)len / r / 2.0;
}
int main() {
  scanf("%I64d%I64d", &a, &b);
  if (b > a || b == 0)
    printf("-1\n");
  else if (a == b) {
    printf("%I64d\n", a);
  } else {
    double ans1 = solve(a - b);
    double ans2 = solve(a + b);
    printf("%.10f\n", ((ans1) < (ans2) ? (ans1) : (ans2)));
  }
  return 0;
}
