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 N = 2e5 + 5;
const int M = 2e5 + 5;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
const double eps = 1e-6;
int a, b;
int dcmp(double x) {
  if (fabs(x) < eps) return 0;
  if (x > eps) return 1;
  return -1;
}
int main() {
  cin >> a >> b;
  if (!a && !b) return 0 * puts("0");
  double best = 1e18;
  if (a > b) {
    double v = (a - b + 0.0) / 2;
    int up = v / b, down = 1;
    if (up >= down) best = min(best, v / up);
  }
  double v = (a + b + 0.0) / 2;
  int up = (v / b), down = 1;
  if (up >= down) best = min(best, v / up);
  if (fabs(best - 1e18) < eps)
    puts("-1");
  else
    printf("%.12f", best);
  return 0;
}
