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;
#define rep(i, n) for (int i = 0; i < (int)(n); i++)

typedef long long ll;
typedef pair<int, int> P;

const ll mod =  1000000007;  //ansを出力する時は (ans % mod + mod) % mod とする(負の剰余を正にする)
const int inf = 1e9;
const long long INF = 1LL << 60;   // INFの値は1152921504606846976


int n,c;
const int V_MAX = 505;
ll d[V_MAX][V_MAX];



int main()
{
    cin >> n >> c;
    rep(i,c) rep(j,c) cin >> d[i][j];
    ll g[n][n];
    rep(i,n) {
        rep(j,n){
            cin >> g[i][j];
            g[i][j]--;
        }
    }
    ll ans = INF;
    ll mod0[c],mod1[c],mod2[c];
    rep(i,c){
        mod0[i] = 0;
        mod1[i] = 0;
        mod2[i] = 0;
    }
    rep(i,c){
        rep(y,n){
            rep(x,n){
                if((x+y) % 3 == 0) mod0[i] += d[g[y][x]][i];
                else if((x+y) % 3 == 1) mod1[i] += d[g[y][x]][i];
                else mod2[i] += d[g[y][x]][i];
            }
        }
    }
    for(int i = 0; i < c; i++){
        for(int j = 0; j < c; j++){
            if(i == j) continue;
            for(int k = 0; k < c; k++){
                if(k == j || i == k) continue;
                ans = min(ans,mod0[i] + mod1[j] + mod2[k]);
            }
        }
    }
    cout << ans << endl;
}