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 C++14 (GCC 5.4.1) (judge verdict: Accepted)

#include <cstdio>
#include <iostream>
#include <cmath>
#include <algorithm>
#include <vector>
#include <queue>
#include <map>
 
using namespace std;

int main(){
	
	string col = "RGBY";
	
	int H, W, d;
	cin >> H >> W >> d;
	
	if(d % 2 != 0){
		for(int i = 0; i < H; i++){
			for(int j = 0; j < W; j++){
				cout << col[(i + j) % 4];
			}
			cout << endl;
		}
		cout << endl;
		return 0;
	}
	
	string hoge[2];
	hoge[0] = "RG";
	hoge[1] = "BY";
	
	for(int i = 10000; i < 10000 + H; i++){
		for(int j = 0; j < W; j++){
			int dum = j;
			if((i + j) % 2 != 0){
				dum++;
			}
			cout << hoge[((i - dum) / d) % 2][((i + dum) / d) % 2];
		}
		cout << endl;
	}
	cout << endl;
	
	return 0;
}