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: Wrong Answer)

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

#define REP(i,s,n) for(int i=s; i<n; ++i)
#define rep(i,n) REP(i,0,n)
#define SORT(c) sort((c).begin(),(c).end())
#define IINF INT_MAX
#define LLINF LLONG_MAX
#define DEBUG false

typedef long long ll;
typedef pair<int, int> ii;

string s = "RGBY";
vector<string> s2 = {"RGGR", "BYYB", "BYYB", "RGGR"};


int main(){

  int h, w, d;
  cin >> h >> w >> d;

  if(d % 2 == 1){
    rep(i, h){
      rep(j, w) cout << s[(i + j) % 4];
      cout << endl;
    }

  }
  else{
    if(d == 2){
      rep(i, h){
        rep(j, w) cout << s[2*(i%2)+((j + 2 * ((i/2) % 2))%4)/2];
        cout << endl;
      }
    }
    else{
      srand((unsigned) time(NULL));
      rep(i, h){
        rep(j, w) cout << s[rand() % 4];
        cout << endl;
      }
    }
  }

  return 0;
}