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: Accepted)

#pragma GCC optimize("Ofast")
#pragma GCC target("sse,sse2,sse3,ssse3,sse4,popcnt,abm,mmx,avx")
#pragma GCC optimize("unroll-loops")
//#pragma warning(disable : 4996)

#ifdef _MSC_VER
#include <intrin.h>
#define __builtin_popcount __popcnt
#define __builtin_popcountll __popcnt64
#endif

#include <limits.h>
#include <math.h>
#include <time.h>

#include <algorithm>
#include <array>
#include <bitset>
#include <cassert>
#include <complex>
#include <cstdio>
#include <cstring>
#include <functional>
#include <iomanip>
#include <iostream>
#include <iterator>
#include <map>
#include <numeric>
#include <queue>
#include <random>
#include <set>
#include <stack>
#include <string>
#include <unordered_map>
#include <unordered_set>
#include <vector>

using namespace std;

#define REP(i, n) for (int i = 0; i < (n); ++i)
#define REPR(i, n) for (int i = n - 1; i >= 0; --i)
#define FOR(i, m, n) for (int i = m; i < n; ++i)
#define FORR(i, m, n) for (int i = m - 1; i >= n; --i)
#define SORT(v, n) sort(v, v + n);
#define VSORT(v) sort(v.begin(), v.end());
#define REVERSE(v, n) reverse(v, v + n);
#define VREVERSE(v) reverse(v.begin(), v.end())
#define ll long long
#define print(x) cout << (x) << '\n'
#define pe(x) cout << (x) << " "
#define DEBUG(x) cout << #x << ": " << x << endl
#define lb(v, n) lower_bound(v.begin(), v.end(), (n))
#define ub(v, n) upper_bound(v.begin(), v.end(), (n))
#define int long long
#define double long double
#define all(x) (x).begin(), (x).end()
#define print_space(v) REP(i, v.size()) cout << v[i] << ((i == v.size() - 1) ? "\n" : " ")
template <typename T1, typename T2> inline void chmin(T1& a, T2 b) {if (a > b) a = b;}
template <typename T1, typename T2> inline void chmax(T1& a, T2 b) {if (a < b) a = b;}
void fail() { cout << -1 << endl; exit(0); }

typedef pair<int, int> pii;
typedef pair<long long, long long> pll;
std::random_device rd;
std::mt19937 mt(rd());
constexpr ll MOD = 998244353;
constexpr int MAX = 100050;
const double pi = acos(-1);
constexpr double EPS = 1e-8;
constexpr ll LINF = 1e18 + 1;
constexpr int INF = 1e9 + 1;
int dx[4] = { 0,0,-1,1 };
int dy[4] = { 1,-1,0,0 };

long long fac[MAX], finv[MAX], inv[MAX];

// テーブルを作る前処理
void COMinit() {
	fac[0] = fac[1] = 1;
	finv[0] = finv[1] = 1;
	inv[1] = 1;
	for (int i = 2; i < MAX; i++) {
		fac[i] = fac[i - 1] * i % MOD;
		inv[i] = MOD - inv[MOD%i] * (MOD / i) % MOD;
		finv[i] = finv[i - 1] * inv[i] % MOD;
	}
}

// 二項係数計算
long long COM(int n, int k) {
	if (n < k) return 0;
	if (n < 0 || k < 0) return 0;
	return fac[n] * (finv[k] * finv[n - k] % MOD) % MOD;
}
ll add(ll x, ll y) {
	x += y;
	if (x >= MOD) return x - MOD;
	return x;
}
ll sub(ll x, ll y) {
	x -= y;
	if (x < 0) return x + MOD;
	return x;
}
ll mult(ll x, ll y) {
	return (x * y) % MOD;
}
ll bin_pow(ll x, ll p) {
	if (p == 0) return 1;
	if (p & 1) return mult(x, bin_pow(x, p - 1));
	return bin_pow(mult(x, x), p / 2);
}

//const int mod = 1000000007;
const int mod = 998244353;
struct mint {
	ll x; // typedef long long ll;
	mint(ll x = 0) :x((x%mod + mod) % mod) {}
	mint operator-() const { return mint(-x); }
	mint& operator+=(const mint a) {
		if ((x += a.x) >= mod) x -= mod;
		return *this;
	}
	mint& operator-=(const mint a) {
		if ((x += mod - a.x) >= mod) x -= mod;
		return *this;
	}
	mint& operator*=(const mint a) { (x *= a.x) %= mod; return *this; }
	mint operator+(const mint a) const { return mint(*this) += a; }
	mint operator-(const mint a) const { return mint(*this) -= a; }
	mint operator*(const mint a) const { return mint(*this) *= a; }
	mint pow(ll t) const {
		if (!t) return 1;
		mint a = pow(t >> 1);
		a *= a;
		if (t & 1) a *= *this;
		return a;
	}

	// for prime mod
	mint inv() const { return pow(mod - 2); }
	mint& operator/=(const mint a) { return *this *= a.inv(); }
	mint operator/(const mint a) const { return mint(*this) /= a; }
};
istream& operator>>(istream& is, const mint& a) { return is >> a.x; }
ostream& operator<<(ostream& os, const mint& a) { return os << a.x; }

char color[4] = {
	'R','Y','G','B'
};

char A[1010][1010];

void solve() {
	int H, W, d; cin >> H >> W >> d;
	REP(i, 1000)REP(j, 1000)A[i][j] = '.';
	if (d & 1) {
		REP(i, 1000) {
			REP(j, 1000) {
				if ((i + j) & 1)A[i][j] = color[0];
				else A[i][j] = color[1];
			}
		}
		REP(i, H) {
			REP(j, W) {
				cout << A[i][j];
			}
			cout << endl;
		}
		return;
	}
	int cnt = 0;
	FOR(si, -1000, 1000) {
		int sj = -si;
		int x = si, y = sj;
		//01
		if (cnt%d < d / 2) {
			while (x<1000||y<1000) {
				REP(_, d / 2) {
					if (x >= 0 && x < 1000 && y >= 0 && y < 1000)A[x][y] = color[0];
					x++; y++;
				}
				REP(_, d / 2) {
					if (x >= 0 && x < 1000 && y >= 0 && y < 1000)A[x][y] = color[1];
					x++; y++;
				}
			}
		}
		//23
		else {
			while (x<1000 || y<1000) {
				REP(_, d / 2) {
					if (x >= 0 && x < 1000 && y >= 0 && y < 1000)A[x][y] = color[2];
					x++; y++;
				}
				REP(_, d / 2) {
					if (x >= 0 && x < 1000 && y >= 0 && y < 1000)A[x][y] = color[3];
					x++; y++;
				}
			}
		}
		cnt++;
	}
	REP(i, 1000) {
		FOR(j, 1, 1000) {
			if (A[i][j] == '.')A[i][j] = A[i][j - 1];
		}
	}
	//REP(i, 10) {
	//	REP(j, 10) {
	//		cout << A[i][j];
	//	}cout << endl;
	//}
	FOR(i, 1, H + 1) {
		FOR(j, 1, W + 1) {
			cout << A[i][j];
		}cout << endl;
	}
} 





signed main() {
	cin.tie(0);
	ios::sync_with_stdio(false);
	//mt.seed((int)(time(0)));
	//int q;
	//cin >> q;
	//while (q--)
	solve();
}