A polygon is lowered at a constant speed of $v$ metres per minute from the air into a liquid that dissolves it at a constant speed of $c$ metres per minute from all sides. Given a point $(x,y)$ inside the polygon that moves with the polygon, determine when the liquid reaches the point.
The border between air and liquid always has y-coordinate 0, and the liquid only eats away from the sides of the polygon in 2 dimensions. The polygon does not rotate as it is lowered into the liquid, and at time 0, it is not touching the liquid.
Unlike the polygon, which is flat (2-dimensional), the liquid exists in three dimensions. Therefore, the liquid seeps into cavities in the polygon. For example, if the polygon is “cup-shaped”, the liquid can get “inside” the cup, as in the diagram below.
The input consists of several test cases.
The first line of each test case contains the five integers $N$, $x$, $y$, $v$, and $c$, where $3 \leq N \leq 30$, $-100 \leq x \leq 100$, $1 \leq y \leq 100$, and $1 \leq c < v \leq 10$.
The following $N$ lines of the test case each contain one vertex of the polygon. The $i$-th line contains the two integers $x$, $y$, where $-100 \leq x \leq 100$, $1 \leq y \leq 100$.
The vertices of the polygon are given in counter-clockwise order. The border of the polygon does not intersect or touch itself, and the point $(x,y)$ lies strictly inside the polygon—it does not lie on the border of the polygon.
Input is terminated by a line containing 0 0 0 0 0. These zeros are not a test case and should not be processed.
For each test case, output the first time in minutes that the liquid reaches the specified point, rounded to four decimal places.
|Sample Input 1||Sample Output 1|
4 0 50 2 1 -1 10 1 10 1 90 -1 90 0 0 0 0 0