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;
const int maxn = 805;
int main() {
  int a, b;
  scanf("%d%d", &a, &b);
  double x;
  double ans = 0.0;
  if (a - b > 0) {
    double temp = (double)(a - b);
    for (int k = (int)(temp / 2.0 / b); k >= 1; k--) {
      ans = temp / 2.0 / (double)k;
      break;
    }
  }
  double temp = (double)(a + b);
  for (int k = (int)(temp / 2.0 / b); k >= 1; k--) {
    if (fabs(ans) > 1e-6)
      ans = min(ans, temp / 2.0 / (double)k);
    else
      ans = temp / 2.0 / (double)k;
    break;
  }
  if (fabs(ans) > 1e-6)
    printf("%.12lf", ans);
  else
    printf("-1\n");
}
