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 9.2.1) (judge verdict: Runtime Error)

#include <bits/stdc++.h>
using namespace std;

int h, w, d;
string col = "RGBY";
vector<string> res;
vector<vector<int>> memo;

void solve();

int main() {
  cin >> h >> w >> d;
  solve();
  for (auto s : res) cout << s << endl;
  return 0;
}

void solve() {
  memo.assign(h, vector<int>(w, 0));
  res.resize(h);
  if (d & 1) {
    for (int i = 0; i < h; ++i)
      for (int j = 0; j < w; ++j) memo[i][j] = (i + j) & 1;
  } else if (d == 2) {
    for (int i = 0; i < h; ++i)
      for (int j = 0; j < w; ++j)
        memo[i][j] = (i & 1) * 2 + (((j / 2) ^ (i / 2)) & 1);
  } else
    assert(0);
  for (int i = 0; i < h; ++i)
    for (int j = 0; j < w; ++j) res[i] += col[memo[i][j]];
}