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<iostream>
#include<cstdio>
#include<cstdlib>
#include<algorithm>
using namespace std;
int n,c;
int w[50][50];
int w2[3][50];
int s[3][300000];
int t[3];
int ans[3];
int flag[100];
int x;


int check(int x,int y)
{
    int tans=0;
    for (int i=0; i<t[x]; ++i)
        tans+=w[s[x][i]][y];
    return tans;
}

int dfs(int now)
{
    if (now==3) return 0;
    int ans=2000000000;
    for (int i=1; i<=c; ++i)
        if (flag[i]==0)
        {
            flag[i]=1;
            ans=min(ans,w2[now][i]+dfs(now+1));
            flag[i]=0;
        }
    return ans;
}
int main()
{
    cin>>n>>c;
    for (int i=1; i<=c; ++i)
        for (int j=1; j<=c; ++j)
            scanf("%d",&w[i][j]);
    for (int i=1; i<=n; ++i)
        for (int j=1; j<=n; ++j)
        {
            scanf("%d",&x);
            s[(i+j)%3][t[(i+j)%3]++]=x;
        }
    for (int i=0; i<3; ++i)
    {
        for (int j=1; j<=c; ++j)
            w2[i][j]=check(i,j);
    }
    cout<<dfs(0)<<endl;
    return 0;
}