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.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.Scanner;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 *
 * @author mikit
 */
public class Main {
    public static void main(String[] args) {
        InputStream inputStream = System.in;
        OutputStream outputStream = System.out;
        Scanner in = new Scanner(inputStream);
        PrintWriter out = new PrintWriter(outputStream);
        DGoodGrid solver = new DGoodGrid();
        solver.solve(1, in, out);
        out.close();
    }

    static class DGoodGrid {
        public void solve(int testNumber, Scanner in, PrintWriter out) {
            int n = in.nextInt(), c = in.nextInt();
            int[][] count = {new int[c], new int[c], new int[c]};
            int[][] d = new int[c][];
            for (int i = 0; i < c; i++) {
                d[i] = new int[c];
            }
            for (int i = 0; i < c; i++) {
                for (int j = 0; j < c; j++) {
                    d[j][i] = in.nextInt();
                }
            }
            for (int i = 1; i <= n; i++) {
                for (int j = 1; j <= n; j++) {
                    count[(i + j) % 3][in.nextInt() - 1]++;
                }
            }
            int[][] cost = {new int[c], new int[c], new int[c]};
            for (int i = 0; i < c; i++) {
                cost[0][i] = dot(count[0], d[i]);
                cost[1][i] = dot(count[1], d[i]);
                cost[2][i] = dot(count[2], d[i]);
            }
            int res = Integer.MAX_VALUE;
            for (int i = 0; i < c; i++) {
                for (int j = 0; j < c; j++) {
                    if (i == j) {
                        continue;
                    }
                    for (int k = 0; k < c; k++) {
                        if (i == k || j == k) {
                            continue;
                        }
                        res = Math.min(cost[0][i] + cost[1][j] + cost[2][k], res);
                    }
                }
            }
            out.println(res);
        }

        static int dot(int[] a1, int[] a2) {
            int r = 0;
            for (int i = 0; i < a1.length; i++) {
                r += a1[i] * a2[i];
            }
            return r;
        }

    }
}