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"
#define rep(i,n) for(int i = 0; i < n; i++)
typedef long long ll;
using namespace std;
#define llMAX numeric_limits<long long>::max()
#define intMAX numeric_limits<int>::max()
#define d_5 100000
#define d9_7 1000000007
#define vll vector<vector<long long>>
#define vl vector<long long>
#define vi vector<int>
#define vii vector<vector<int>>
#define pb push_back
#define pf push_front
int main(void){
  int n,C;
  cin>>n>>C;
  vii d=vii(C,vi(C));
  vii c=vii(n,vi(n));
  rep(i,C){
    rep(j,C){
      cin>>d[i][j];
    }
  }
  rep(i,n){
    rep(j,n){
      cin>>c[i][j];
      c[i][j]--;
    }
  }
  vii modcolor(3,vi());
  rep(i,n){
    rep(j,n){
      modcolor[(i+j)%3].pb(c[i][j]);
    }
  }
  /*sort(modcolor[0].begin(),modcolor[0].end());
  sort(modcolor[1].begin(),modcolor[1].end());
  sort(modcolor[2].begin(),modcolor[2].end());*/
  //cout<<"e";
  vii cost=vii(3,vi(C));
  int tmp=0;
  rep(i,C){
    tmp=0;
    rep(i2,modcolor[0].size()){
      tmp+=d[modcolor[0][i2]][i];
    }
    cost[0][i]=tmp;
  }
  rep(i,C){
    tmp=0;
    rep(i2,modcolor[1].size()){
      tmp+=d[modcolor[1][i2]][i];
    }
    cost[1][i]=tmp;
  }
  rep(i,C){
    tmp=0;
    rep(i2,modcolor[2].size()){
      tmp+=d[modcolor[2][i2]][i];
    }
    cost[2][i]=tmp;
  }
  int ans=intMAX;
  rep(i,C){
    rep(j,C){
      rep(k,C){
        if(i==j || j==k || k==i)continue;
        ans=min(ans,cost[0][i]+cost[1][j]+cost[2][k]);
      }
    }
  }
  cout<<ans<<endl;
  return 0;
}