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 static java.lang.Math.*;
import java.math.BigInteger;

public class Main {
	public static void main(String[] args){
		Scanner sc = new Scanner(System.in);
		
		// 入力
		int n = sc.nextInt();
		int c = sc.nextInt();
		int[][] gridD = new int[c+1][c+1];
		int[][] gridC = new int[n+1][n+1];
		for(int i = 1; i < c+1; i++){
		    for(int j = 1; j < c+1; j++){
		        gridD[i][j] = sc.nextInt();
		    }
		}
		for(int i = 1; i < n+1; i++){
		    for(int j = 1; j < n+1; j++){
		        gridC[i][j] = sc.nextInt();
		    }
		}
		
		// 計算
		int result = Integer.MAX_VALUE;
		int[][] d0 = new int[c+1][2];
		d0[0][1] = Integer.MAX_VALUE;
		d0[1][0] = c+1;
		int[][] d1 = new int[c+1][2];
		d1[0][1] = Integer.MAX_VALUE;
		d1[1][0] = c+2;
		int[][] d2 = new int[c+1][2];
		d2[0][1] = Integer.MAX_VALUE;
		d2[1][0] = c+3;
		for(int k = 1; k < c+1; k++){
		    for(int i = 1; i < n+1; i++){
		        for(int j = 1; j < n+1; j++){
		            if((i+j)%3 == 0){
		                d0[k][0] = k;
		                d0[k][1] += gridD[gridC[i][j]][k];
		            }else if((i+j)%3 == 1){
		                d1[k][0] = k;
		                d1[k][1] += gridD[gridC[i][j]][k];
		            }else{
		                d2[k][0] = k;
		                d2[k][1] += gridD[gridC[i][j]][k];
		            }
		        }
		    }
		}
		Arrays.sort(d0, (s, t) -> Integer.compare(s[1], t[1]));
		Arrays.sort(d1, (s, t) -> Integer.compare(s[1], t[1]));
		Arrays.sort(d2, (s, t) -> Integer.compare(s[1], t[1]));
		for(int i = 0; i < c; i++){
		    for(int j = 0; j < c; j++){
		        for(int k = 0; k < c; k++){
		            if(d0[i][0] == d1[j][0] || d1[j][0] == d2[k][0] || d2[k][0] == d0[i][0]) continue;
		            result = min(result, d0[i][1] + d1[j][1] + d2[k][1]);
		        }
		    }
		}
		// 出力
		System.out.println(result);
	}
}