Problem:
Score : 400 points

Problem Statement

There is a grid with N rows and N columns of squares. Let (i,j) be the square at the i-th row from the top and the j-th column from the left.

These squares have to be painted in one of the C colors from Color 1 to Color C. Initially, (i,j) is painted in Color c_{i,j}.

We say the grid is a good grid when the following condition is met for all i,j,x,y satisfying 1 \leq i,j,x,y \leq N:

If (i+j) \% 3=(x+y) \% 3, the color of (i,j) and the color of (x,y) are the same.

If (i+j) \% 3 \neq (x+y) \% 3, the color of (i,j) and the color of (x,y) are different.

Here, X \% Y represents X modulo Y.

We will repaint zero or more squares so that the grid will be a good grid.

For a square, the wrongness when the color of the square is X before repainting and Y after repainting, is D_{X,Y}.

Find the minimum possible sum of the wrongness of all the squares.

Constraints

1 \leq N \leq 500

3 \leq C \leq 30

1 \leq D_{i,j} \leq 1000 (i \neq j),D_{i,j}=0 (i=j)

1 \leq c_{i,j} \leq C

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N C
D_{1,1} ... D_{1,C}
:
D_{C,1} ... D_{C,C}
c_{1,1} ... c_{1,N}
:
c_{N,1} ... c_{N,N}

Output

If the minimum possible sum of the wrongness of all the squares is x, print x.

Sample Input 1

2 3
0 1 1
1 0 1
1 4 0
1 2
3 3

Sample Output 1

3

Repaint (1,1) to Color 2. The wrongness of (1,1) becomes D_{1,2}=1.

Repaint (1,2) to Color 3. The wrongness of (1,2) becomes D_{2,3}=1.

Repaint (2,2) to Color 1. The wrongness of (2,2) becomes D_{3,1}=1.

In this case, the sum of the wrongness of all the squares is 3.

Note that D_{i,j} \neq D_{j,i} is possible.

Sample Input 2

4 3
0 12 71
81 0 53
14 92 0
1 1 2 1
2 1 1 2
2 2 1 3
1 1 2 2

Sample Output 2

428

Solution in Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.*;
import java.lang.*;

class Counter<T> extends TreeMap<T,Long>{
    public Counter(){
        super();
    }
    public Long count(Object elm){
        return getOrDefault(elm,0L);
    }
    public void add(T elm, long amount){
        if(!this.containsKey(elm)) put(elm, amount);
        else replace(elm, get(elm)+amount);
    }
}

public class Main {
    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        int N = sc.nextInt();
        int C = sc.nextInt();
        int[][] D = new int[C][C];
        for(int x=0; x<C; x++) for(int y=0; y<C; y++) D[x][y]=sc.nextInt();

        int[][] grid = new int[N][N];
        for(int r=0; r<N; r++) for(int c=0; c<N; c++) grid[r][c]=sc.nextInt()-1;

        Counter<Integer>[] colorCount = new Counter[3];
        for(int p=0;p<3;p++) colorCount[p]=new Counter<>();
        for(int r=0; r<N; r++) for(int c=0; c<N; c++) colorCount[(r+c)%3].add(grid[r][c],1);
        
        int ans = Integer.MAX_VALUE;
        for(int c0=0; c0<C; c0++) for(int c1=0; c1<C; c1++) for(int c2=0; c2<C; c2++){
            if(c0!=c1 && c1!=c2 && c2!=c0){
                int wrongness = 0;
                for(int c=0; c<C; c++){
                    wrongness += D[c][c0] * colorCount[0].count(c);
                    wrongness += D[c][c1] * colorCount[1].count(c);
                    wrongness += D[c][c2] * colorCount[2].count(c);
                }
                ans = Math.min(ans, wrongness);
            }
        }
        System.out.println(ans);


    }
}