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 9.2.1) (judge verdict: Wrong Answer)

#include<stdio.h>

int main(void){
  int n , c;
  scanf("%d %d",&n ,&c);

  int d[c+1][c+1];
  for(int i = 1;i <= c;i ++){
    for(int j = 1;j <= c;j ++){
      scanf("%d",&d[i][j]);
    }
  }

  int k[n+1][n+1];
  int waru1[31] = {0} , waru2[31] = {0} , waru0[31] = {0};
  for(int i = 1;i <= n;i ++){
    for(int j = 1;j <= n;j ++){
      scanf("%d",&k[i][j]);
      if((i+j)%3 == 1){
        waru1[k[i][j]] ++;
      }
      else if((i+j)%3 == 2){
        waru2[k[i][j]] ++;
      }
      else{
        waru0[k[i][j]] ++;
      }
    }
  }

  /*while(flag == 1){
    flag = 0;
    for(int i = 1;i <= c;i ++){
      for(int j = 1;j <= c;j ++){
        for(int k = 1;k <= c;k ++){
          if(d[i][j] > d[i][k] + d[k][j]){
            d[i][j] = d[i][k] + d[k][j];
            flag = 1;
          }
        }
      }
    }
  }*/

  int min = 1000000000;
  for(int i = 1;i <= c;i ++){
    for(int j = 1;j <= c;j ++){
      for(int x = 1;x <= c;x ++){
        if(i != j && j != x && x != i){
          int tas = 0;
          for(int y = 1;y <= c;y ++){
            tas += waru1[y] * d[y][i];
            tas += waru2[y] * d[y][j];
            tas += waru0[y] * d[y][x];
          }
          
          if(min > tas && waru1[i] + waru2[j] + waru0[x] < n*n){
            min = tas;
          }
        }
      }
    }
  }

  printf("%d\n",min);

  return 0;
}