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;
void Read(int &x) {
  char c;
  while (c = getchar(), c != EOF)
    if (c >= '0' && c <= '9') {
      x = c - '0';
      while (c = getchar(), c >= '0' && c <= '9') x = x * 10 + c - '0';
      ungetc(c, stdin);
      return;
    }
}
int a, b;
double ans1 = -1, ans2 = -1;
int main() {
  Read(a), Read(b);
  if (a == b) {
    printf("%.12lf\n", (double)b);
    return 0;
  }
  if (a < b) {
    printf("-1");
    return 0;
  }
  int t = a - b, c;
  c = t / b;
  if (c & 1) c--;
  while (1.0 * t / c >= b) c += 2;
  c -= 2;
  if (1.0 * t / c >= b) ans1 = 1.0 * t / c;
  t = a + b;
  c = t / b;
  if (c & 1) c--;
  while (1.0 * t / c >= b) c += 2;
  c -= 2;
  if (1.0 * t / c >= b) ans2 = 1.0 * t / c;
  if (ans1 == -1 && ans2 == -1) {
    printf("-1");
    return 0;
  } else if (ans2 == -1 || ans1 == -1)
    ans2 = max(ans1, ans2);
  else
    ans2 = min(ans1, ans2);
  printf("%.12lf", ans2);
}
