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>

int main() {
	int h, w, d, color;
	short c[510][510] = {};
	scanf("%d%d%d", &h, &w, &d);

	for (int i = 0; i < h; i++)
	{
		for (int j = 0; j < w; j++)
		{
			color = c[i][j];
			for (int k = 0; k < d; k++)
			{
				if (i + d - k < h && j + k < w
					&& c[i + d - k][j + k] == color) {
					c[i + d - k][j + k] = (c[i + d - k][j + k] + 1) % 4;
				}
				if (i + d - k < h && j - k >= 0 
					&& c[i + d - k][j - k] == color) {
					c[i + d - k][j - k] = (c[i + d - k][j - k] + 1) % 4;
				}
			}
			if (j + d < w
				&& c[i][j + d] == color) {
				c[i][j + d] = (c[i][j + d] + 1) % 4;
			}
		}
	}

	for (int i = 0; i < h; i++)
	{
		for (int j = 0; j < w; j++)
		{
			switch (c[i][j])
			{
			case 0:
				printf("R");
				break;
			case 1:
				printf("Y");
				break;
			case 2:
				printf("G");
				break;
			default:
				printf("B");
				break;
			}

		}
		printf("\n");
	}

	return 0;
}