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 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>

int ans[505][505];
int main(){
	int i, j, h, w, d;
	char color[6] = "RGBY";

	scanf("%d%d%d", &h, &w, &d);

	int z, x, y;
	if (d % 2 == 1) {
		z = (d+1)/2;
		for (i = 0; i < h; i++) {
			for (j = 0; j < w; j++) {
				x = i / z;
				y = j / z;
				ans[i][j] = (x+y)%2;
			}
		}
	} else {
		z = d / 2 + 2;
		for (i = 0; i < h; i++) {
			for (j = 0; j < w; j++) {
				x = i / (z-1);
				y = j / (z-2);
				if (x%2==0 && y%2==0)ans[i][j] = 0;
				else if(x%2==0 && y%2==1)ans[i][j] = 1;
				else if(x%2==1 && y%2==0)ans[i][j] = 2;
				else ans[i][j]=3;
			}
		}
	}
	
	for (i = 0; i < h; i++) {
		for (j = 0; j < w; j++){
			printf("%c", color[ans[i][j]]);
		}
		printf("\n");
	}

	return 0;
}