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.io.BufferedReader;
import java.io.InputStreamReader;
import java.util.Scanner;
 
public class Main {
 
    public static void main(String[] args) {
        Scanner scan = new Scanner(new BufferedReader(new InputStreamReader(System.in)));
        int n = scan.nextInt();
        int c = scan.nextInt();
        int[][] d = new int[c][c];
        for (int i = 0; i < c; i++) {
            for (int j = 0; j < c; j++) {
                d[i][j] = scan.nextInt();
            }
        }
        int[][] g = new int[n][n];
        for (int i = 0; i < n; i++) {
            for (int j = 0; j < n; j++) {
                g[i][j] = scan.nextInt() - 1;
            }
        }
        Main main = new Main();
        System.out.println(main.solve(n, c, d, g));
    }

    public int solve(int n, int c, int[][] d, int[][] g) {
        int[][] count_of_colors = new int[3][30];
        for(int i = 0; i < n; i++) {
            for(int j = 0; j < n; j++) {
                count_of_colors[(i+j)%3][g[i][j]] += 1;
            }
        }
        int answer = Integer.MAX_VALUE;
        for (int i = 0; i < c; i++) {
            int first_group_difference = difference(d, c, count_of_colors[0], i);
            for (int j = 0; j < c; j++) {
                if (i == j) {
                    continue;
                }
                int second_group_difference = difference(d, c, count_of_colors[1], j);
                for (int k = 0; k < c; k++) {
                    if (i == k || j == k) {
                        continue;
                    }
                    int third_group_difference = difference(d, c, count_of_colors[2], k);
                    answer = Math.min(answer, first_group_difference + second_group_difference + third_group_difference);
                }
            }
        }
        return answer;
    }

    public int difference(int[][] d, int c, int[] count, int color) {
        int diff = 0;
        for(int i = 0; i < c; i++) {
            diff += count[i] * d[i][color];
        }
        return diff;
    }
 
}