Problem: 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}

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}

Examples

Input

2 2 1


Output

RY
GR


Input

2 3 2


Output

RYB
RGB

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
typedef long long i64; typedef vector<int> ivec; typedef vector<string> svec;
int H,W,d;

int main(int argc, char *argv[])
{
  scanf("%d%d%d",&H,&W,&d);
  int stride = (W/d+1)*d;
  
  for (int i = 0; i < H; i++) {
    for (int j = 0; j < W; j++) {
      int x,y;
      x = stride+i+j;y = stride+i-j;
      if((x+1)/d%2==1&&(y+1)/d%2==1) printf("R");
      else if((x+1)/d%2==0&&(y+1)/d%2==1) printf("G");
      else if((x+1)/d%2==0&&(y+1)/d%2==0) printf("B");
      else if((x+1)/d%2==1&&(y+1)/d%2==0) printf("Y");
    }
    printf("\n");
  }
  
  return 0;
}