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: Wrong Answer)

#include <bits/stdc++.h>

#define endl '\n'
#define max_n 2002

using namespace std;

int side;
int co[max_n][max_n];

void dfs(int x, int y, int val) {
	if(x < 0 || y < 0 || x >= 2002 || y >= 2002) return;
	if(co[x][y] != 0) return;
	for(int i=x-side; i<=x+side; i++) {
		for(int j=y-side; j<=y+side; j++) {
			if(i < 0 || j < 0 || i >= 2002 || j >= 2002) continue;
			if(abs(x-i) + abs(y-j) <= side) co[i][j] = val;
		}
	}
	dfs(x + side + side + 1, y + side + side - 1, val);
	dfs(x - side - side - 1, y - side - side + 1, val);
	dfs(x + side + side - 1, y - side - side - 1, val);
	dfs(x - side - side + 1, y - side - side + 1, val);
}

int main() {
	ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
	int n, m, d;
	cin >> n >> m >> d;
	side = (d-1) / 2;
	memset(co, 0, sizeof co);
	dfs(1000, 1000, 1);
	dfs(1000+side+1, 1000+side, 2);
	dfs(1000+side, 1000-side-1, 3);
	dfs(1000+side+side+1, 1000-1, 4);
	for(int i=0; i<n; i++) {
		for(int j=0; j<m; j++) {
			if(co[1000+i][1000+j] == 1) {
				cout << 'R';
			}
			else if(co[1000+i][1000+j] == 2) {
				cout << 'G';
			}
			else if(co[1000+i][1000+j] == 3) {
				cout << 'Y';
			}
			else if(co[1000+i][1000+j] == 4) {
				cout << 'B';
			}
		}
		cout << endl;
	}
	return 0;
}