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 main(void){
	int N,C;
	int i,j,ans;
	scanf("%d %d",&N,&C);
	int D[C][C],color[3][30];
	for(i=0;i<C;i++){
		for(j=0;j<C;j++){
			scanf("%d",&D[i][j]);
		}
	}
	int x;
	for(i=0;i<3;i++){
		for(j=0;j<30;j++){
			color[i][j]=0;
		}
	}
	for(i=0;i<N;i++){
		for(j=0;j<N;j++){
			scanf("%d",&x);
			color[(i+j)%3][x-1]++;// %3で割ったチームにx色が何個あるか
		}
	}
		
	ans = 1e9;int k,l;
	for(i=0;i<C;i++){
		for(j=0;j<C;j++){
			if(i != j){
				for(k=0;k<C;k++){
					if(i != k && j != k){
						int a = 0;
						for(l=0;l<C;l++){
							a += D[l][i]*color[0][l]; //%3=0チームでl色からi色に塗り替え
							a += D[l][j]*color[1][l]; //%3=1チームでl色からj色に塗り替え
							a += D[l][k]*color[2][l]; //%3=2チームでl色からk色に塗り替え
						}
						if(a < ans) ans = a;
					}
				}
			}
		}
	}
	
	
	printf("%d",ans);
	return 0;}