CF645G.Armistice Area Apportionment

After a drawn-out mooclear arms race, Farmer John and the Mischievous Mess Makers have finally agreed to establish peace. They plan to divide the territory of Bovinia with a line passing through at least two of the $n$ outposts scattered throughout the land. These outposts, remnants of the conflict, are located at the points $(x_{1},y_{1}),(x_{2},y_{2}),...,(x_{n},y_{n})$ .

In order to find the optimal dividing line, Farmer John and Elsie have plotted a map of Bovinia on the coordinate plane. Farmer John's farm and the Mischievous Mess Makers' base are located at the points $P=(a,0)$ and $Q=(-a,0)$ , respectively. Because they seek a lasting peace, Farmer John and Elsie would like to minimize the maximum difference between the distances from any point on the line to $P$ and $Q$ .

Formally, define the difference of a line relative to two points $P$ and $Q$ as the smallest real number $d$ so that for all points $X$ on line , $|PX-QX|<=d$ . (It is guaranteed that $d$ exists and is unique.) They wish to find the line passing through two distinct outposts $(x_{i},y_{i})$ and $(x_{j},y_{j})$ such that the difference of relative to $P$ and $Q$ is minimized.

The first line of the input contains two integers $n$ and $a$ ( $2<=n<=100000$ , $1<=a<=10000$ ) — the number of outposts and the coordinates of the farm and the base, respectively.

The following $n$ lines describe the locations of the outposts as pairs of integers $(x_{i},y_{i})$ ( $|x_{i}|,|y_{i}|<=10000$ ). These points are distinct from each other as well as from $P$ and $Q$ .

Print a single real number—the difference of the optimal dividing line. Your answer will be considered correct if its absolute or relative error does not exceed $10^{-6}$ .

Namely: let's assume that your answer is $a$ , and the answer of the jury is $b$ . The checker program will consider your answer correct, if .

In the first sample case, the only possible line is $y=x-1$ . It can be shown that the point $X$ which maximizes $|PX-QX|$ is $(13,12)$ , with , which is .

In the second sample case, if we pick the points $(0,1)$ and $(0,-3)$ , we get as $x=0$ . Because $PX=QX$ on this line, the minimum possible difference is $0$ .