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: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;

#define int long long
#define all(v) (v).begin(), (v).end()
#define rall(v) (v).rbegin(), (v).rend()
#define rep(i,n) for(int i=0;i<n;++i)
#define rep1(i,n) for(int i=1;i<n;++i)
#define exrep(i, a, b) for(ll i = a; i < b; i++)
#define out(x) cout << x << endl
#define EPS (1e-7)
#define gearup ios_base::sync_with_stdio(0);cin.tie(0);cout.tie(0);

typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
typedef vector<int> vi;
typedef vector<char> vc;
typedef vector<bool> vb;
typedef vector<double> vd;
typedef vector<string> vs;
typedef vector<pair<int,int> > vpii;
typedef vector<vector<int> > vvi;
typedef vector<vector<char> > vvc;
typedef vector<vector<bool> > vvb;
typedef vector<vector<double> > vvd;
typedef vector<vector<string> > vvs;
typedef vector<ll> vl;
typedef vector<vector<ll> > vvl;
typedef vector<vector<vector<ll> > > vvvl;
ll MOD = 1000000007;
const long long L_INF = 1LL << 60;
const int INF = 2147483647; // 2^31-1
const double PI = acos(-1);
//cout<<fixed<<setprecision(10);

template<class T> inline bool chmin(T& a, T b) {if (a > b) {a = b;return true;}return false;}
template<class T> inline bool chmax(T& a, T b) {if (a < b) {a = b;return true;}return false;}
template<class T> void debug(T v){rep(i,v.size()) cout<<v[i]<<" ";cout<<endl;}
const ll dx[8] = {1, 1, 0, -1, -1, -1, 0, 1};
const ll dy[8] = {0, 1, 1, 1, 0, -1, -1, -1};
//以降 cin の入力元が 'input.txt' になる
//std::ifstream in("input.txt");
//std::cin.rdbuf(in.rdbuf());

signed main()
{
    gearup;
    ll n,C; cin >> n >> C;
    vvl d(C,vl(C));
    rep(i,C)rep(j,C)cin>>d[i][j];
    vvl c(n,vl(n));
    rep(i,n)rep(j,n)cin>>c[i][j];
    ll res = L_INF;
    rep(i,C){ //あまり0でぬる色
        rep(j,C){ //1
            rep(k,C){ //2
                if(i == j || j == k || k == i)continue;
                ll ans = 0;
                rep(ii,n)rep(jj,n){
                    int num = ((ii+1) + (jj+1)) % 3;
                    if(num == 0){
                        if(c[ii][jj] == i+1)continue;
                        ans += d[c[ii][jj] - 1][i];
                    }
                    else if(num == 1){
                        if(c[ii][jj] == j+1)continue;
                        ans += d[c[ii][jj] - 1][j];
                    }
                    else{
                        if(c[ii][jj] == k+1)continue;
                        ans += d[c[ii][jj] - 1][k];
                    }
                }
                res = min(ans,res);
            }
        }
    }
    out(res);
}