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<bits/stdc++.h>

using namespace std;
using ll = long long;
using ull = unsigned long long;
using P = pair<int,int>;
using Pl = pair<ll,ll>;
#define REP(i,n) for(ll i = 0;i < (ll)n;i++)
#define REPD(i,n) for(ll i = (ll)n - 1;i >= 0;i--)
#define ALL(x) (x).begin(),(x).end()
#define FILLl(n,first_dimension_size,value) fill((ll*)n,(ll*)(n+first_dimension_size),value)
#define FILL(n,first_dimension_size,value) fill((int*)n,(int*)(n+first_dimension_size),value)
#define INF (1 << 30)
#define MOD 1000000007
//int dy[4] = {-1,0,0,1};
//int dx[4] = {0,1,-1,0};


int main(){

  int n,C;
  cin >> n >> C;
  vector<vector<int>> D(C,vector<int>(C));
  REP(i,C)REP(j,C)cin >> D[i][j];
  vector<vector<int>> c(n,vector<int>(n));
  REP(i,n)REP(j,n)cin >> c[i][j];
  vector<vector<int>> v(3,vector<int>(C,0));
  REP(i,n){
    REP(j,n){
      v[(i+j)%3][c[i][j]-1]++;
    }
  }
  ll res = INF;
  REP(i,C){
    ll d1 = 0;
    REP(l,C)d1 += D[l][i]*v[0][l];
    REP(j,C){
      if(i == j)continue;
      ll d2 = 0;
      REP(l,C)d2 += D[l][j]*v[1][l];
      REP(k,C){
        if(i == k || j == k)continue;
        ll d3 = 0;
        REP(l,C)d3 += D[l][k]*v[2][l];
        res = min(res,d1+d2+d3);
      }
    }
  }
  cout << res << endl;

  

  return 0;
}