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<bits/stdc++.h>
using namespace std;
///Welcome to Nasif's Code
#define bug printf("bug\n");
#define bug2(var) cout<<#var<<" "<<var<<endl;
#define co(q) cout<<q<<endl;
typedef long long int ll;
typedef unsigned long long int ull;
const int MOD = (int)1e9+7;
const int MAX = 1e6;
#define pi acos(-1)
#define inf 1000000000000000LL
#define FastRead    ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
int cost[50][50],grid[505][505],n,c;
ll box[3][50],dp[50][3];
vector<pair<int,int> >v[4];
ll recur(int pos,int taken,int used[])
{
    if(taken==3)
        return 0;
    if(pos>c)
        return INT_MAX;
    ll &ret=dp[pos][taken];
    ret=INT_MAX*3;
    for(int i=0; i<3; i++)
    {
        if(!used[i])
        {
            used[i]=1;
            ret=min(recur(pos+1,taken+1,used)+box[i][pos],ret);
            used[i]=0;
        }
    }
    return ret;
}
int main()
{
    FastRead
    //freopen("output.txt", "w", stdout);
    cin>>n>>c;
    for(int i=1; i<=c; i++)
        for(int j=1; j<=c; j++)
            cin>>cost[i][j];
    for(int i=1; i<=n; i++)
        for(int j=1; j<=n; j++)
            cin>>grid[i][j];
    for(int i=1; i<=n; i++)
    {
        for(int j=1; j<=n; j++)
        {
            int m=(i+j)%3;
            v[m].push_back({i,j});
        }
    }
    int mn=INT_MAX,sz0=v[0].size(),sz1=v[1].size(),sz2=v[2].size();
    for(int i=1; i<=c; i++)
    {
        ll cur=0;
        for(int j=0; j<sz0; j++)
        {
            int x=v[0][j].first;
            int y=v[0][j].second;
            if(grid[x][y]!=i)
            {
                cur+=cost[grid[x][y]][i];
            }
        }
        box[0][i]=cur;

        cur=0;
        for(int j=0; j<sz1; j++)
        {
            int x=v[1][j].first;
            int y=v[1][j].second;
            if(grid[x][y]!=i)
            {
                cur+=cost[grid[x][y]][i];
            }
        }
        box[1][i]=cur;
        cur=0;
        for(int j=0; j<sz2; j++)
        {
            int x=v[2][j].first;
            int y=v[2][j].second;
            if(grid[x][y]!=i)
            {
                cur+=cost[grid[x][y]][i];
            }
        }
        box[2][i]=cur;
    }
   /* for(int i=0;i<3;i++){
        for(int j=1;j<=c;j++)
            cout<<box[i][j]<<" ";
        cout<<endl;
    }*/
    int used[5]= {0};
    cout<<recur(1,0,used)<<endl;


    return 0;
}