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: Wrong Answer)

import java.util.*;

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 ][ C ];
    for(int i = 0 ;i < C ; i ++){
      for(int j = 0 ; j < C ; j ++){
        D[i][j] = sc.nextInt();
      }
    }

    int[][] c = new int[ N ][ N ];
    int[][] D_3 = new int [ 3 ] [ C ] ;
    for(int i = 0 ; i < N ; i ++){
      for( int j = 0 ; j < N ; j ++ ){
        c[ i ][ j ] = sc.nextInt();
        for(int k = 0 ; k < C ; k ++){
          D_3[ ( i + j ) % 3 ][ k ] += D[ c[i][j] - 1 ][ k ] ;
        }
      }
    }

    long DMin = 25000000L;
    for(int i = 0 ; i < C ; i ++){
      for(int j = 0 ; j < C ;j ++){
        for(int k = 0 ; k < C ; k ++){
          if( i != j && j != k && k != i ){
            long DSum = D_3[0][i] + D_3[1][j] + D_3[2][k];
            if( DSum < DMin ){
              DMin = DSum ;
            }
          }
        }
      }
    }

    System.out.println( DMin );

  }
}