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 double eps = 0.0000000001;
int sgn(double f) {
  if (fabs(f) < eps) return 0;
  return f > 0 ? 1 : -1;
}
double a, b;
int main() {
  scanf("%lf%lf", &a, &b);
  long long d;
  if (a == b) {
    printf("%.9lf\n", a);
    return 0;
  }
  if (b - a > 0) {
    printf("-1\n");
    return 0;
  }
  bool f = false;
  double ans;
  double dc = (a - b) / (2 * b);
  d = dc;
  if (sgn(0.0 + d + 1 - dc) == 0) {
    d = d + 1;
  }
  double x1 = 100000000;
  if (d != 0) {
    ans = max((a - b) / (2.0 * d), b);
    f = true;
  }
  dc = (a + b) / (2 * b);
  d = dc;
  if (sgn(0.0 + d + 1 - dc) == 0) {
    d = d + 1;
  }
  if (b != 0)
    ;
  {
    if (f)
      ans = min(ans, max((a + b) / (2.0 * d), b));
    else
      ans = max((a + b) / (2.0 * d), b);
  }
  printf("%.9lf\n", ans);
  return 0;
}