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>
#include<stdlib.h>

#define POS(i,j) ((i)*c+(j))

void run(void){
  int n,c;
  scanf("%d%d",&n,&c);
  int *d=(int *)malloc(sizeof(int)*c*c);
  int i,j;
  for(i=0;i<c*c;i++) scanf("%d",d+i);
  int *cnt=(int *)calloc(3*c,sizeof(int));
  for(i=0;i<n;i++){
    for(j=0;j<n;j++){
      int m;
      scanf("%d",&m);
      m--;
      cnt[POS((i+j)%3,m)]++;
    }
  }
  int min=1000*n*n;
  int x,y,z;
  for(x=0;x<c;x++){
    for(y=0;y<c;y++){
      for(z=0;z<c;z++){
	if(x==y || y==z || z==x) continue;
	int sum=0;
	for(i=0;i<c;i++){
	  sum+=cnt[POS(0,i)]*d[POS(i,x)];
	  sum+=cnt[POS(1,i)]*d[POS(i,y)];
	  sum+=cnt[POS(2,i)]*d[POS(i,z)];
	}
	if(min>sum) min=sum;
      }
    }
  }
  printf("%d\n",min);
}

int main(void){
  run();
  return 0;
}