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 Java8 (OpenJDK 1.8.0) (judge verdict: Accepted)

import java.util.Scanner;

class Main{
	public static void main(String[] args) {
		Scanner sc=new Scanner(System.in);
		int N=sc.nextInt();
		int C=sc.nextInt();
		int[][] iwakan=new int[C][C];
		for(int i=0; i<C; i++) {
			for(int j=0; j<C; j++) {
				iwakan[i][j]=sc.nextInt();
			}
		}
		int[] mod0_color=new int[C];
		int[] mod1_color=new int[C];
		int[] mod2_color=new int[C];
		for(int i=0; i<C; i++) {
			mod0_color[i]=0;
			mod1_color[i]=0;
			mod2_color[i]=0;
		}
		for(int i=0; i<N; i++) {
			for(int j=0; j<N; j++) {
				if((i+j)%3==0) {
					mod0_color[sc.nextInt()-1]++;
				}
				else if((i+j)%3==1) {
					mod1_color[sc.nextInt()-1]++;
				}
				else if((i+j)%3==2) {
					mod2_color[sc.nextInt()-1]++;
				}
			}
		}
		//mod 0 基準
		long sum_iwa=Long.MAX_VALUE;
		long tmp_iwa=0;
		for(int i=0; i<C; i++) {		//mod 0の色
			for(int j=0; j<C; j++) {	//mod 1の色
				for(int k=0; k<C; k++) {//mod 2の色
					tmp_iwa=0;
					if(i!=j && j!=k && k!=i) {
						for(int l=0; l<C; l++) {
							if(i!=l) {
								tmp_iwa+=mod0_color[l]*iwakan[l][i];
							}
						}
						for(int l=0; l<C; l++) {
							if(j!=l) {
								tmp_iwa+=mod1_color[l]*iwakan[l][j];
							}
						}
						for(int l=0; l<C; l++) {
							if(k!=l) {
								tmp_iwa+=mod2_color[l]*iwakan[l][k];
							}
						}
						sum_iwa=Math.min(tmp_iwa, sum_iwa);
					}
				}
			}
		}
		System.out.println(sum_iwa);
	}
}