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: Time Limit Exceeded)

#include<bits/stdc++.h> 
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define ForkD(i,k,n) for(int i=n;i>=k;i--)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  
#define Lson (o<<1)
#define Rson ((o<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,0x3f,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define MEMx(a,b) memset(a,b,sizeof(a));
#define INF (0x3f3f3f3f)
#define F (1000000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int> 
#define pi pair<int,int>
#define SI(a) ((a).size())
#define Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;
#define PRi2D(a,n,m) For(i,n) { \
						For(j,m-1) cout<<a[i][j]<<' ';\
						cout<<a[i][m]<<endl; \
						} 
#pragma comment(linker, "/STACK:102400000,102400000")
#define ALL(x) (x).begin(),(x).end()
#define gmax(a,b) a=max(a,b);
#define gmin(a,b) a=min(a,b);
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return ((a-b)%F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
inline int read()
{
	int x=0,f=1; char ch=getchar();
	while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
	while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
	return x*f;
} 
int n,m,d;
int a[1010][1010];
int Abs(int i,int j) {
	return max(i-j,j-i);
}
char h[1000]="RYGB";
int main()
{
//	freopen("data.in","r",stdin);
//	freopen("D.out","w",stdout);
	
	int n=read(),m=read(),d=read();
	
		int lsz=0,i;
		if (d%2==1) {
			For(i,n) For(j,m) a[i][j]=(i+j)&1;
			For(i,n) {
				For(j,m) printf("%c",h[a[i][j]]);puts("");
			}
		} else {
			int pm=m;
			m=max(2*n,max(d,m));
				for(i=1;lsz<=n;i++) {
					For(j,m) {
						bool b[1000]={};
						For(k,i) {
							int j2;
							if(k==i) {
								j2=j-d;
								if (0<j2&&j2<=m)
									b[a[k][j2]]=1;
							}else{
								j2=j-(d-(i-k)); 
								if (0<j2&&j2<=m) b[a[k][j2]]=1;
								j2=j+(d-(i-k)); 
								if (0<j2&&j2<=m)b[a[k][j2]]=1;
									
							}
							
						}
						int tp=0;
						while(b[tp]) ++tp;
						a[i][j]=tp;
						if (tp>=4) lsz=-1;
					}
					++lsz;
				}	
				int n2=i-1,l1=n2-n+1;
				Fork(i,l1,n2) {
					For(j,pm) printf("%c",h[a[i][j]]);puts("");
				}
			}
		//	cout<<lsz<<' '<<i<<endl;
			
	
	
	
	return 0;
}