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;
using namespace std::chrono;
const long double EPS = 1e-8;
const long double PI = acos(-1.0);
void run() {
  int a, b;
  scanf("%d%d", &a, &b);
  if (a - b < 0)
    printf("-1\n");
  else if (a == b)
    printf("%d\n", a);
  else {
    int k = (a - b) / (2 * b);
    double ans = (a - b) / (2.0 * k);
    k = (a + b) / (2 * b) - 1;
    if (k >= 0) ans = min(ans, double(a + b) / (2 * k + 2));
    printf("%.15lf\n", ans);
  }
}
int main() {
  run();
  return 0;
}