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 <bits/stdc++.h>
#define loop(n) for (int ngtkana_is_genius = 0; ngtkana_is_genius < int(n); ngtkana_is_genius++)
#define rep(i, begin, end) for(int i = int(begin); i < int(end); i++)
#define all(v) v.begin(), v.end()
#define lint long long
auto cmn = [](auto& a, auto b){if (a > b) {a = b; return true;} return false;};
auto cmx = [](auto& a, auto b){if (a < b) {a = b; return true;} return false;};

template <typename T>
std::istream& operator>> (std::istream& is, std::vector<T>& v) {
  for (auto & x : v) is >> x;
  return is;
}

template <typename T>
std::ostream& operator<< (std::ostream& os, const std::vector<T>& v) {
  auto n = v.size();
  os << "{";
  for (size_t i = 0; i < n; i++)
    {os << (i > 0 ? "," : "") << v.at(i);}
  return os << "}";
}

void debug_impl() { std::cerr << std::endl; }
template <typename Head, typename... Tail>
void debug_impl(Head head, Tail... tail){
  std::cerr << " " << head;
  debug_impl(tail...);
}
#define debug(...)\
  std::cerr << std::boolalpha << "[" << #__VA_ARGS__ << "]:";\
  debug_impl(__VA_ARGS__);\
  std::cerr << std::noboolalpha;

template <typename T>
auto make_vector_impl(size_t sz, T t) {return std::vector<T>(sz, t);}

template <size_t N, typename T, typename U, std::enable_if_t<N == 1, std::nullptr_t> = nullptr>
auto make_vector(size_t sz, U u) {return make_vector_impl(sz, T(u));}

template <size_t N, typename T, std::enable_if_t<N == 1, std::nullptr_t> = nullptr>
auto make_vector(size_t sz) {return std::vector<T>(sz);}

template <size_t N, typename T, typename... Args, std::enable_if_t<N != 1, std::nullptr_t> = nullptr>
auto make_vector(size_t a, Args... args) {return make_vector_impl(a, make_vector<N - 1, T>(args...));}

template <typename T, typename Size_t>
auto& at(T& t, Size_t i) {return t.at(i);}

template <typename T, typename Size_t, typename... Args>
auto& at(T& t, Size_t i, Args... args) {return at(t.at(i), args...);}

int main() {
  std::cin.tie(0); std::cin.sync_with_stdio(false);
  int h, w, d; std::cin >> h >> w >> d;
  auto a = make_vector< 2, int >(h, w, -1);
  rep(i, 0, h) rep(j, 0, w) {
    at(a, i, j) =
        (((i + j    ) / d) % 2) * 1
      + (((i + w - j) / d) % 2) * 2;
  }
  for(auto& v : a) {
    for (auto x : v) {
      std::cout << (x == 0
        ? 'R'
        : x == 1
        ? 'Y'
        : x == 2
        ? 'G'
        : 'B'
        );
    }
    std::cout << std::endl;
  }
  return 0;
}