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>
using namespace std;
int main(){
  int N,C_;cin>>N>>C_;
  vector<vector<int>> D(C_,vector<int>(C_));
  vector<vector<int>> c(N,vector<int>(N));
  for(int i=0;i<C_;i++){
    for(int j=0;j<C_;j++)cin>>D[i][j];//iは行
  }
  for(int i=0;i<N;i++){
    for(int j=0;j<N;j++)cin>>c[i][j];//iは行
  }
  vector<int> A;
  vector<int> B;
  vector<int> C;
  for(int i=0;i<N;i++){
    for(int j=0;j<N;j++){
      if((i+j)%3==0)A.push_back(c[i][j]-1);
      if((i+j)%3==1)B.push_back(c[i][j]-1);
      if((i+j)%3==2)C.push_back(c[i][j]-1);
    }
  }
  int ans=INT_MAX;
  for(int ca=0;ca<C_;ca++){
    for(int cb=0;cb<C_;cb++){
      for(int cc=0;cc<C_;cc++){
        if(ca==cb||cb==cc||cc==ca)continue;
        int tmp=0;
        for(int i=0;i<A.size();i++)tmp+=D[A[i]][ca];
        for(int i=0;i<B.size();i++)tmp+=D[B[i]][cb];
        for(int i=0;i<C.size();i++)tmp+=D[C[i]][cc];
        if(tmp<ans)ans=tmp;
      }
    }
  }
  cout<<ans;
}