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++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#pragma warning (disable:4996)
#include<stdio.h>
#include<iostream>
#include<vector>
using namespace std;
int main()
{
	int n, see;
	cin >> n>>see;
	int d[40][40] = { 0 }, c[505][505] = { 0 };
	for (int i = 1; i <= see; i++)
	{
		for (int j = 1; j <= see; j++)
		{
			cin >> d[i][j];
		}
	}
	for (int i = 1; i <= n; i++)
	{
		for (int j = 1; j <= n; j++)
		{
			cin >> c[i][j];
		}
	}
	int zdai = 99999999;

	for (int k = 1; k <= see; k++)
	{
		int dai0 = 0;//余0颜色的代价
		for (int i = 1; i <= n; i++)
		{
			for (int j = 1; j <= n; j++)
			{
				if ((i + j) % 3 == 0) dai0 = dai0 + d[c[i][j]][k];
			}
		}

		for (int l = 1; l <= see; l++)
		{
			if (l == k) continue;
			int dai1 = 0;//余1颜色的代价
			for (int i = 1; i <= n; i++)
			{
				for (int j = 1; j <= n; j++)
				{
					if ((i + j) % 3 == 1) dai1 = dai1 + d[c[i][j]][l];
				}
			}

			for (int m = 1; m <= see; m++)//格子每种颜色组合
			{
				if (m == k || m == l) continue;

				int dai2 = 0;//余2颜色的代价
				for (int i = 1; i <= n; i++)
				{
					for (int j = 1; j <= n; j++)
					{
						if ((i + j) % 3 == 2) dai2 = dai2 + d[c[i][j]] [m];
					}
				}
				if (dai0+dai1+dai2 < zdai) zdai = dai0 + dai1 + dai2;
			}
		}
	}
	//printf("%d", zdai);
	//int zdai0 = 999999999;
	//for (int k = 1; k <= see; k++)//余0的格子每种颜色
	//{
	//	if (dai0 < zdai0) zdai0 = dai0;
	//}

	//int zdai1 = 999999999;
	//for (int k = 1; k <= see; k++)//余1的格子每种颜色
	//{
	//	if (dai1 < zdai1) zdai1 = dai1;
	//}

	//int zdai = zdai0 + zdai1 + zdai2;
	



	//for (int sum = 2; sum <= 2*n; sum++)//每个斜行
	//{
	//	if (sum <= n + 1) {
	//		for (int i = 1; i <= sum - 1; i++)//每个格子
	//		{
	//			for (int j = 1; j <= see; j++)//每种颜色
	//			{

	//			}
	//		}}
	//	else {
	//		for (int i = sum-n; i <= n; i++)
	//		{

	//		}}
	//}
	printf("%d", zdai);
	return 0;
}