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 print(int a, int b) {
  printf("%d.", a / b);
  a = a % b * 10;
  for (int i = 1; i <= 20; i++) {
    printf("%d", a / b);
    a = a % b * 10;
  }
}
int main() {
  int a, b;
  int temp1, temp2;
  double t1, t2;
  scanf("%d%d", &a, &b);
  temp1 = (a + b) / (b * 2);
  if (temp1 < 1)
    t1 = -1;
  else
    t1 = (a + b) * 1.0 / (temp1 * 2.0);
  temp2 = (a - b) / (b * 2);
  if (temp2 > 0)
    t2 = (a - b) * 1.0 / (temp2 * 2.0);
  else
    t2 = -1;
  if (t1 != -1 && (t1 < t2 || t2 == -1))
    print(a + b, temp1 * 2);
  else if (t2 != -1)
    print(a - b, temp2 * 2);
  else
    printf("-1\n");
}