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: Time Limit Exceeded)

#include <bits/stdc++.h>
#include <math.h>
#define _GLIBCXX_DEBUG
#define _LIBCPP_DEBUG 0

using namespace std;
#define ll long long
#define rep(i,n) for (int i = 0; i < n; i++)
#define rrep(i,n) for (int i = n-1; i >= 0; i--)
#define MOD (1000000007)
#define vi vector<int>
#define vl vector<ll>
#define vb vector<bool>
#define vvi vector<vi>
#define vvl vector<vl>
#define pii pair<int, int>
#define pli pair<ll, int>a
#define pb push_back
#define sz(x) int(x.size())

#define mp make_pair
#define all(a) (a).begin(),(a).end()

template<class T> inline bool chmax(T& a, T b) { if (a < b) { a = b; return true; } return false; }
template<class T> inline bool chmin(T& a, T b) { if (a > b) { a = b; return true; } return false; }

ll gcd(ll a, ll b) {
    if (b == 0) return a;
    else return gcd(b, a % b);
}

ll keta(ll n){
  string s = to_string(n);
  ll num = s.size();
  return num;
}

const ll INF = 1LL << 60;

const int dh[4] = {1,0,-1,0};
const int dw[4] = {0,1,0,-1};

struct Edge{
  int to;
  int weight;
  Edge(int t, int w) : to(t), weight(w){}
};
using Graph = vector<vector<Edge>>;
using P = pair<ll, int>;

class UnionFind{
  public:
  vi Parent;
  UnionFind(int n){
    Parent = vi(n,-1);
  }

  int root(int a){
    if(Parent[a] < 0) return a;
    else return Parent[a] = root(Parent[a]);
  }

  int size(int a){
    return -Parent[root(a)];
  }

  bool merge(int a, int b){
    a = root(a);
    b = root(b);
    if(a == b) return false;

    if(size(a) < size(b)) swap(a,b);
    Parent[a] += Parent[b];
    Parent[b] = a;

    return true;
  }
};

class Factrial{
  public:
  vl Fac;
  Factrial(int MAX){
    Fac = vl(MAX+1);
    rep(i,MAX) {
      if(i == 0) Fac[i+1] = 1;
      else Fac[i+1] = ((i+1)*Fac[i])%MOD;
    }
  }
};

int main(){
  int N,C; cin >> N >> C;
  vvi D(C,vi(C));
  rep(i, C) rep(j, C) cin >> D[i][j];

  vvi c(N, vi(N));
  rep(i,N) rep(j,N){
    cin >> c[i][j];
    c[i][j]--;
  }

  ll ans = INF;
  vvi G(3);
  rep(i,N) rep(j,N) {
    G[(i+j)%3].pb(c[i][j]);
  }


  rep(i,C) rep(j,C) {
    if(i == j) continue;
    rep(k,C) {
      if(i == k || j == k) continue;
      // 色をi,j,kの三色で統一した場合
      ll sum = 0;
      for(auto X : G[0]){
        sum += D[X][i];
        if(sum >= ans) goto next;
      } 
      for(auto X : G[1]){
        sum += D[X][j];
        if(sum >= ans) goto next;
      } 
      for(auto X : G[2]){
        sum += D[X][k];
        if(sum >= ans) goto next;
      } 
      next:
      chmin(ans,sum);

    }
  }

  cout << ans << endl;
}