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 C# (Mono 4.6.2.0) (judge verdict: Accepted)

using System;
using System.Text;
using System.Collections.Generic;
using static System.Math;
using System.Linq;
using static System.Array;

namespace Vertial
{
    class Program
    {
        static int[][] D;
        static int[][] map;
        static int N;
        static int[,] dist;
        static int C;
        static void Main(string[] args)
        {
            var NC = Console.ReadLine().Split(' ').Select(x => int.Parse(x)).ToArray();
            N = NC[0];
            C = NC[1];

            D = new int[C][];
            for (int i = 0; i < C; i++)
            {
                D[i] = Console.ReadLine().Split(' ').Select(x => int.Parse(x)).ToArray();
            }

            map = new int[N][];
            for (int i = 0; i < N; i++)
            {
                map[i] = Console.ReadLine().Split(' ').Select(x => int.Parse(x)).ToArray();
            }
            dist = new int[3,C+1];
            for (int i = 1; i <= N; i++)
            {
                for (int j = 1; j <= N; j++)
                {
                    if ((i + j) % 3 == 0)
                    {
                        dist[0, map[i - 1][j - 1]]++;
                    }
                    else if ((i + j) % 3 == 1)
                    {
                        dist[1, map[i - 1][j - 1]]++;
                    }
                    else
                    {
                        dist[2, map[i - 1][j - 1]]++;
                    }
                }
            }
            var min = long.MaxValue;
            
            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 || i== k))
                        {
                            //3割って0余るものをi,1あまるものをj,2余るものをに変換する
                            min = Min(min, Calculation(i, j, k));
                        }
                    }
                }
            }
            Console.WriteLine(min);
        }

        static long Calculation(int a,int b,int c)
        {
            long ans = 0;
            for (int i = 0; i < 3; i++)
            {
                for (int j = 1; j <= C; j++)
                {
                    if(i == 0)
                    {
                        ans += dist[i, j] * D[j - 1][a-1];
                    }else if(i == 1)
                    {
                        ans += dist[i, j] * D[j - 1][b-1];
                    }
                    else
                    {
                        ans += dist[i, j] * D[j - 1][c-1];
                    }
                    
                }
            }
            
            return ans;
        }


    }
}