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;
const int INF = 0x3f3f3f3f;
void Open() {}
const double eps = 1e-11;
int dcmp(double x) {
  if (x < eps) return -1;
  return x > eps;
}
int main() {
  int a, b;
  scanf("%d%d", &a, &b);
  if (a < b) {
    printf("-1\n");
    return 0;
  }
  int len = a + b;
  double low = len * 1.0 / 2.0 / b;
  int tmp = floor(low);
  double ans1 = -1;
  if (tmp != 0) ans1 = len * 1.0 / 2.0 / tmp;
  len = a - b;
  double ans2 = -1;
  low = len * 1.0 / 2.0 / b;
  tmp = floor(low);
  if (tmp != 0) ans2 = len * 1.0 / 2.0 / tmp;
  if (dcmp(ans1) < 0 && dcmp(ans2) < 0) {
    printf("-1\n");
    return 0;
  }
  if (dcmp(ans1) < 0) {
    printf("%.10f\n", ans2);
    return 0;
  }
  if (dcmp(ans2) < 0) {
    printf("%.10f\n", ans1);
    return 0;
  }
  printf("%.10f\n", min(ans1, ans2));
  return 0;
}