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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<utility>
using namespace std;
#define MS(m,z) memset(m,z,sizeof(m))
typedef unsigned U;typedef long long ll;typedef pair<int,int > P;
void inp();
int main(){
#ifdef XS
	//freopen(".in","r",stdin);
#endif
	//int ttt;scanf("%d",&ttt);for(int i=1;i<=ttt;i++)
	inp();
	return 0;
}
int G[600][600];
int dp[5][50];
int C[50][50];
int ans = 1e9;
int p[3];
int n,x,c;
void jd(){
	int cur = 0;
	for(int i = 0;i<3;i++){
		for(int j = 1;j<=c;j++){
			cur += dp[i][j] * C[ j ][ p[ i] ];
		}
	}
	ans = min(ans, cur);
}
void inp(){
	scanf("%d", &n); scanf("%d",&c);
	for(int i = 1;i<=c;i++){
		for(int j =1;j<=c;j++){
			scanf("%d", &C[i][j]); // from i to j
		}
	}
	for(int i =1;i<=n;i++){
		for(int j=1;j<=n;j++){
			scanf("%d",&G[i][j]);
			int r = (i+j)%3;
			dp[r][ G[i][j] ] ++;
		}
	}
	for(int i = 1;i<=c;i++){
		for(int j = 1;j<=c;j++){
			for(int k = 1;k<=c;k++){
				if( i == j || j == k || i == k)continue;
				p[0]=i;
				p[1]=j;
				p[2]=k;
				jd();
			}
		}
	}
	printf("%d", ans);
}
//d.cc by xsthunder at Fri Jul 27 23:58:32 2018