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 f(double z, double first) {
  int k = 0;
  double w;
  while (first > 2 * z) {
    w = z;
    while (2 * w < first) w *= 2, k += 2;
    first -= w;
  }
  if (first <= z)
    return first;
  else
    return 2 * z - first;
}
int main() {
  int b, c;
  double a, lo, hi, mid, w;
  int k;
  while (scanf("%d %d", &c, &b) == 2) {
    a = c;
    if (c < b) {
      printf("-1\n");
    } else if (c == b) {
      printf("%d\n", b);
    } else {
      k = ((int)(floor(a / b + 1e-9) + 1e-9 - 1)) / 2;
      hi = a / (2 * k + 1);
      lo = a / (2 * k + 2);
      k = 60;
      while (k--) {
        mid = (lo + hi) / 2.0;
        w = f(mid, a);
        if (w <= b)
          lo = mid;
        else
          hi = mid;
      }
      printf("%.10lf\n", mid);
    }
  }
  return 0;
}