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 double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
const double eps = 1e-10;
const long long INF = (1LL << 60);
int main() {
  int a, b;
  scanf("%d%d", &a, &b);
  if (a == b) {
    printf("%d\n", a);
    return 0;
  }
  double ans = -1;
  if (a >= b) {
    int k = (int)floor((a * 1.0 / b - 1) / 2);
    ans = (a - b) * 1.0 / (2 * k);
    ans = min(ans, (a + b) * 1.0 / (2 * k + 2));
  }
  printf("%0.12f\n", ans);
  return 0;
}
