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;
import java.util.HashMap;

/**
 * Built using CHelper plug-in Actual solution is at the top
 *
 * @author Silviase
 */
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) {
      // i+jの同値類に分けておいて,色と数をカウントしておく.
      // あとはコストを各色の組(c_0, c_1, c_2)について考えればいいよねっ☆
      // と思ったけど各グリッドについてやっていたらTLEしそう.個数を持っておけば間に合わん？30個しかないやん
      // そり！
      int n = in.nextInt();
      int c = in.nextInt();
      int[][] dif = new int[c + 1][c + 1];
      HashMap<Integer, Integer>[] hm = new HashMap[3];
      hm[0] = new HashMap<>();
      hm[1] = new HashMap<>();
      hm[2] = new HashMap<>();
      int minCost = Integer.MAX_VALUE;

      for (int i = 1; i <= c; i++) {
        for (int j = 1; j <= c; j++) {
          dif[i][j] = in.nextInt();
        }
      }

      for (int i = 1; i <= n; i++) {
        for (int j = 1; j <= n; j++) {
          int color = in.nextInt();
          if (hm[(i + j) % 3].containsKey(color)) {
            // 既に存在するので書換
            hm[(i + j) % 3].replace(color, hm[(i + j) % 3].get(color) + 1);
          } else {
            hm[(i + j) % 3].put(color, 1);
          }
        }
      }

      for (int i = 1; i <= c; i++) {
        for (int j = 1; j <= c; j++) {
          for (int k = 1; k <= c; k++) {
            if (i == j || j == k || k == i) {
              continue;
            } else {
              // 書換がいる.
              int tmpCost = 0;
              for (int l = 1; l <= c; l++) {
                if (hm[0].containsKey(l)) {
                  tmpCost += hm[0].get(l) * dif[l][i];
                }
                if (hm[1].containsKey(l)) {
                  tmpCost += hm[1].get(l) * dif[l][j];
                }
                if (hm[2].containsKey(l)) {
                  tmpCost += hm[2].get(l) * dif[l][k];
                }
              }
              minCost = Math.min(minCost, tmpCost);
            }
          }
        }
      }

      out.println(minCost);

    }

  }
}