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>
int Min2(int a,int b){
    return a<b?a:b;
}

int main(){
    int n,c;
    scanf("%d %d",&n,&c);
    int d[30][30];
    int cnt[3][30]={0};
    int i,j,k;
    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++){
            int tmp;
            scanf("%d",&tmp);
            cnt[(i+j)%3][tmp-1]++;
        }
    }
/*
    puts("");
    for(i=0;i<3;i++){
        for(j=0;j<c;j++){
            printf("%d ",cnt[i][j]);
        }
        puts("");
    }
*/
    int iwa[3][30]={0};
    for(i=0;i<3;i++){
        for(j=0;j<c;j++){
            for(k=0;k<c;k++){
                iwa[i][j]+=d[k][j]*cnt[i][k];
            }
        }
    }
/*
    puts("");
    for(i=0;i<3;i++){
        for(j=0;j<c;j++){
            printf("%d ",iwa[i][j]);
        }
        puts("");
    }
*/
    int ans=1000000000;
    for(i=0;i<c;i++){
        for(j=0;j<c;j++){
            if(i==j)continue;
            for(k=0;k<c;k++){
                if(i==k||j==k)continue;
                //printf("%d %d %d:%d %d %d %d\n",i,j,k,ans,iwa[0][i],iwa[1][j],iwa[k]);
                ans=Min2(ans,iwa[0][i]+iwa[1][j]+iwa[2][k]);
                //printf("%d %d %d:%d\n",i,j,k,ans);
            }
        }
    }
    printf("%d\n",ans);
}