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;

public class Main {

	public static void main(String[] args) {
		Scanner sc = new Scanner(System.in);
		int n = sc.nextInt();
		int c = sc.nextInt();
		int[][] d = new int[c + 1][c + 1];
		int[][] grid = new int[3][c + 1];
		for(int i = 1; i < c + 1; i++) {
			for(int j = 1; j < c + 1; j++) {
				d[i][j] = sc.nextInt();
			}
		}
		for(int i = 0; i < 3; i++) {
			for(int j = 0; j < c + 1; j++) {
				grid[i][j] = 0;
			}
		}
		for(int i = 1; i < n + 1; i++) {
			for(int j = 1; j < n + 1; j++) {
				int tmp = sc.nextInt();
				grid[(i + j) % 3][tmp]++;
			}
		}
		int ans = Integer.MAX_VALUE;
		for(int i = 1; i < c + 1; i++) {
			for(int j = 1; j < c + 1; j++) {
				if(i != j) {
					for(int k = 1; k < c + 1; k++) {
						if(i != k && j != k) {
							int dif = 0;
							for(int l = 1; l < c + 1; l++) {
								dif += d[l][i] * grid[0][l];
								dif += d[l][j] * grid[1][l];
								dif += d[l][k] * grid[2][l];
							}
							ans = Math.min(ans, dif);
						}
					}
				}
			}
		}
		System.out.println(ans);

	}

}