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;
void read(int &digit) {
  digit = 0;
  char c;
  for (c = getchar(); (c < '0' || c > '9') && c != '-'; c = getchar())
    ;
  bool type = false;
  if (c == '-') type = true, c = getchar();
  for (; c >= '0' && c <= '9'; digit = digit * 10 + c - '0', c = getchar())
    ;
  if (type == true) digit = -digit;
}
int a, b;
int main() {
  read(a), read(b);
  if (a < b) {
    printf("-1\n");
    return 0;
  }
  if (a == b) {
    printf("%d\n", b);
    return 0;
  }
  a -= b;
  double q, ans, ans1;
  int sum;
  sum = a / b;
  if (sum & 1) sum--;
  if (sum == 0)
    ans = -1;
  else
    ans = (double)a / sum;
  a += 2 * b;
  sum = a / b;
  if (sum & 1) sum--;
  if (sum == 0)
    ans1 = -1;
  else
    ans1 = (double)a / sum;
  if (ans < 0 || (ans1 < ans && ans1 > 0)) ans = ans1;
  if (ans < 0)
    printf("-1\n");
  else
    printf("%.10lf\n", ans);
  return 0;
}