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;
int main() {
  int a, b;
  scanf("%d%d", &a, &b);
  if (b > a)
    puts("-1");
  else if (b == a)
    printf("%d", b);
  else {
    int k = (a - b) / (b << 1);
    double ans = (double)(a - b) / 2 / (k - 1);
    if ((double)(a - b) / 2 / k >= b) ans = (double)(a - b) / 2 / k;
    if ((double)(a - b) / 2 / (k + 1) >= b) ans = (double)(a - b) / 2 / (k + 1);
    k = (a + b) / (b << 1);
    if ((double)(a + b) / 2 / (k - 1) >= b)
      ans = min(ans, (double)(a + b) / 2 / (k - 1));
    if ((double)(a + b) / 2 / k >= b) ans = min(ans, (double)(a + b) / 2 / k);
    if ((double)(a + b) / 2 / (k + 1) >= b)
      ans = min(ans, (double)(a + b) / 2 / (k + 1));
    printf("%.10f", ans);
  }
}