Problem:
Score : 700 points

Problem Statement

We have a grid with H rows and W columns of squares.
We will represent the square at the i-th row from the top and j-th column from the left as (i,\ j).
Also, we will define the distance between the squares (i_1,\ j_1) and (i_2,\ j_2) as |i_1 - i_2| + |j_1 - j_2|.

Snuke is painting each square in red, yellow, green or blue.
Here, for a given positive integer d, he wants to satisfy the following condition:

No two squares with distance exactly d have the same color.

Find a way to paint the squares satisfying the condition.
It can be shown that a solution always exists.

Constraints

2 ≤ H, W ≤ 500

1 ≤ d ≤ H + W - 2

Input

Input is given from Standard Input in the following format:

H W d

Output

Print a way to paint the squares satisfying the condition, in the following format. If the square (i,\ j) is painted in red, yellow, green or blue, c_{ij} should be R, Y, G or B, respectively.

c_{11}c_{12}...c_{1W}
:
c_{H1}c_{H2}...c_{HW}

Sample Input 1

2 2 1

Sample Output 1

RY
GR

There are four pairs of squares with distance exactly 1.
As shown below, no two such squares have the same color.

(1,\ 1), (1,\ 2) : R, Y

(1,\ 2), (2,\ 2) : Y, R

(2,\ 2), (2,\ 1) : R, G

(2,\ 1), (1,\ 1) : G, R

Sample Input 2

2 3 2

Sample Output 2

RYB
RGB

There are six pairs of squares with distance exactly 2.
As shown below, no two such squares have the same color.

(1,\ 1) , (1,\ 3) : R , B

(1,\ 3) , (2,\ 2) : B , G

(2,\ 2) , (1,\ 1) : G , R

(2,\ 1) , (2,\ 3) : R , B

(2,\ 3) , (1,\ 2) : B , Y

(1,\ 2) , (2,\ 1) : Y , R

Solution in Pascal (FPC 2.6.2) (judge verdict: Wrong Answer)

var a:array[-1..501,-1..501,1..4]of boolean;
	b:array[-1..501,-1..501]of integer;
	h,w,d,i,j,k,t,s,be,en:longint;
begin
	readln(h,w,d);
	fillchar(a,sizeof(a),true);
	for i:=1 to h do for j:=1 to w do begin
		k:=1;
		while not a[i,j,k] do inc(k);
		b[i,j]:=k;
		//if j+d<=w then a[i,j+d,k]:=false;
		//if i+d<=h then a[i+d,j,k]:=false;
		be:=i;en:=h;
		if j-d<1 then be:=i+(d-j);
		if i+d<h then en:=i+d;
		for t:=be to en do a[t,j-(d-(t-i)),k]:=false;
		be:=i;
		if w-j<d then be:=i+(d-(w-j));
		for t:=en downto be do a[t,j+(d-(t-i)),k]:=false;
	end;
	for i:=1 to h do begin
		for j:=1 to w do
			case b[i,j] of 
				1:write('R');
				2:write('Y');
				3:write('G');
				4:write('B');
			end;
		writeln;
	end;
end.