Arrow
Contest: Experimental Educational Round: VolBIT Formulas Blitz
Difficulty rating: 2000
Tags: geometry

Petya has recently started working as a programmer in the IT city company that develops computer games.

Besides game mechanics implementation to create a game it is necessary to create tool programs that can be used by game designers to create game levels. Petya's first assignment is to create a tool that allows to paint different arrows on the screen.

A user of this tool will choose a point on the screen, specify a vector (the arrow direction) and vary several parameters to get the required graphical effect. In the first version of the program Petya decided to limit parameters of the arrow by the following: a point with coordinates (px, py), a nonzero vector with coordinates (vx, vy), positive scalars a, b, c, d, a > c.

The produced arrow should have the following properties. The arrow consists of a triangle and a rectangle. The triangle is isosceles with base of length a and altitude of length b perpendicular to the base. The rectangle sides lengths are c and d. Point (px, py) is situated in the middle of the triangle base and in the middle of side of rectangle that has length c. Area of intersection of the triangle and the rectangle is zero. The direction from (px, py) point to the triangle vertex opposite to base containing the point coincides with direction of (vx, vy) vector.

Enumerate the arrow points coordinates in counter-clockwise order starting from the tip.

Input format:
The only line of the input contains eight integers px, py, vx, vy ( - 1000 ≤ px, py, vx, vy ≤ 1000, vx2 + vy2 > 0), a, b, c, d (1 ≤ a, b, c, d ≤ 1000, a > c).

Output format:
Output coordinates of the arrow points in counter-clockwise order. Each line should contain two coordinates, first x, then y. Relative or absolute error should not be greater than 10 - 9.

Constraints:
Time limit: 0.5 s
Memory limit: 64.0 MB

Example input:
8 8 0 2 8 3 4 5

Example output:
8.000000000000 11.000000000000
4.000000000000 8.000000000000
6.000000000000 8.000000000000
6.000000000000 3.000000000000
10.000000000000 3.000000000000
10.000000000000 8.000000000000
12.000000000000 8.000000000000

Editorial:
To get a vector of the given length b in the direction of the given vector (vx, vy) it is just required to normalize the given vector (divide it by its length) and then multiply by b.
Let's denote $$len = \sqrt{vx^2 + vy^2}$$, vnx = vx / len, vny = vy / len. Then (vnx, vny) is the normalized vector, and the first point of the arrow is (px + vnx·b, py + vny·b).
To get the second point of the arrow one needs to rotate the normalized vector 90 degrees counter-clockwise and then multiply by the half of the triangle base a / 2. Let's denote vlx =  - vny, vly = vnx. Then (vlx, vly) is the normalized vector 90 degrees counter-clockwise to (vnx, vny). So the second point of the arrow is (px + vlx·a / 2, py + vly·a / 2).
The third point can be found the same way as the second point but the length of the vector to it is c / 2. So the third point of the arrow is (px + vlx·c / 2, py + vly·c / 2).
The fourth point can be found by adding the vector of the length c / 2 to the left of the given and the vector of the length d reverse to the given. So the fourth point of the arrow is (px + vlx·c / 2 - vnx·d, py + vly·c / 2 - vny·d).
The remaining points are symmetrical to the points discussed above so they can be obtained the same way, just using minus for (vlx, vly) instead of plus. So the next points are (px - vlx·c / 2 - vnx·d, py - vly·c / 2 - vny·d), (px - vlx·c / 2, py - vly·c / 2), (px - vlx·a / 2, py - vly·a / 2).