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: Wrong Answer)

#include <bits/stdc++.h>
using namespace std;
using ll = long long;
//const ll MOD = 1e9 + 7;
//const ll MOD = 998244353;
//const ll MOD = ;
ll mod(ll A, ll M) {return (A % M + M) % M;}
const ll INF = 1LL << 60;
template<class T> bool chmin(T& a, T b) {if (a > b) {a = b; return true;} return false;}
template<class T> bool chmax(T& a, T b) {if (a < b) {a = b; return true;} return false;}
ll divceil(ll A, ll B) {return (A + (B - 1)) / B;}
ll myctoi(char C) {return C - '0';}
char myitoc(ll N) {return '0' + N;}
#define FINALANS(A) {cout << (A) << '\n'; exit(0);}

int main()
{
  ll N, C;
  cin >> N >> C;
  vector<vector<ll>> D(C, vector<ll>(C));
  for (ll i = 0; i < C; i++)
  {
    for (ll j = 0; j < C; j++)
    {
      cin >> D.at(i).at(j);
    }
  }

  vector<vector<ll>> X(C, vector<ll>(3, 0));
  for (ll i = 0; i < N; i++)
  {
    for (ll j = 0; j < N; j++)
    {
      ll c;
      cin >> c;
      c--;
      X.at(c).at((i + j) % 3)++;
    }
  }

  ll ans = INF;
  for (ll i = 0; i < C; i++)
  {
    for (ll j = 0; j < C; j++)
    {
      if (i == j)
        continue;
      
      for (ll k = 0; k < C; k++)
      {
        if (i == k || j == k)
          continue;

        ll tmp = 0;
        for (ll l = 0; l < C; l++)
        {
          tmp += D.at(l).at(i) * X.at(l).at(0);
          tmp += D.at(l).at(j) * X.at(l).at(1);
          tmp += D.at(l).at(k) * X.at(l).at(2);
        }
      }
    }
  }
  cout << ans << endl;
}