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 (GCC 5.4.1) (judge verdict: Accepted)

#include<stdio.h>
#include<stdlib.h>

#define min(a,b) ((a)<(b)?(a):(b))

struct st{
	int num;
	int c;
	int cd;
};

int main(){
	int n,c,i,j,k,l;
	int d[31][31];
	int cc[500][500];
	scanf("%d%d",&n,&c);
	int se[3][100]={0},tt[3]={0};
	for(i=0;i<c;i++){
		for(j=0;j<c;j++){
			scanf("%d",&d[i][j]);
		}
	}
	for(i=0;i<n;i++){
		for(j=0;j<n;j++){
			scanf("%d",&cc[i][j]);
			se[(i+j)%3][cc[i][j]-1]++;
		}
	}
	long long ans=2000000000000000;
	for(i=0;i<c;i++){
		for(j=0;j<c;j++){
			for(k=0;k<c;k++){
				long long tmp=0;
				if(i==j||j==k||k==i)continue;
				for(l=0;l<c;l++)tmp+=d[l][i]*se[0][l];
				for(l=0;l<c;l++)tmp+=d[l][j]*se[1][l];
				for(l=0;l<c;l++)tmp+=d[l][k]*se[2][l];
				ans=min(ans,tmp);
			}
		}
	}
	printf("%lld\n",ans);
	return 0;
}