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;
int main() {
  double a, b;
  scanf("%lf%lf", &a, &b);
  if (b > a) {
    puts("-1");
    return 0;
  }
  if (a == b) {
    printf("%.9f\n", a);
    return 0;
  }
  double an1 = 1e10, an2 = 1e10;
  long long le = 0, ri = 1e9;
  long long mid;
  while (ri >= le) {
    mid = (le + ri) / 2;
    double tmp = (a - b) / 2 / mid;
    if (tmp < b) {
      ri = mid - 1;
    } else {
      le = mid + 1;
      an1 = tmp;
    }
  }
  ri = 1e9;
  le = 0;
  while (ri >= le) {
    mid = (le + ri) / 2;
    double tmp = (a + b) / 2 / mid;
    if (tmp < b) {
      ri = mid - 1;
    } else {
      le = mid + 1;
      an2 = tmp;
    }
  }
  printf("%.9f\n", min(an1, an2));
  return 0;
}
