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;
typedef long long ll;
typedef pair<int, int> P;
#define rep(i,n) for(int i=0;i<n;++i)
#define MOD 1000000007

int main(){
    int n, C;
    cin >> n >> C;

    int d[C][C];
    rep(i, C){
        rep(j, C){
            cin >> d[i][j];
        }
    }

    int c[3][30] = {};
    rep(i, n){
        rep(j, n){
            int tmp;
            cin >> tmp;
            c[(i+j)%3][tmp-1]++;
        }
    }

    int ans = 2002002002;
    rep(i, C){
        rep(j, C){
            if(i!=j){
                rep(k, C){
                    if(i!=k && j!=k){
                        int cnt = 0;
                        rep(l, C){
                            cnt += d[l][i]*c[0][l];
                        }
                        rep(l, C){
                            cnt += d[l][j]*c[1][l];
                        }
                        rep(l, C){
                            cnt += d[l][k]*c[2][l];
                        }

                        ans = min(ans, cnt);
                    }
                }
            }
        }
    }
    cout << ans << endl;
    // printf("%d\n", N);
    return 0;
}