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: Wrong Answer)

#include <bits/stdc++.h>
#define REP(i, n) for(ll i = 0; i < (ll)n; i++)
#define FOR(i, a, b) for(ll i = (a); i < (ll)b; i++)
#define ALL(obj) (obj).begin(), (obj).end()
#define INF 1000000000000000
using namespace std;
typedef long long ll;
typedef double db;
typedef string str;
typedef pair<ll, ll> p;
constexpr int MOD = 1000000007;
template <class T> inline bool chmin(T &a, T b) {
    if(a > b) {
        a = b;
        return true;
    }
    return false;
}
template <class T> inline bool chmax(T &a, T b) {
    if(a < b) {
        a = b;
        return true;
    }
    return false;
}

void print(const std::vector<int> &v) {
    std::for_each(v.begin(), v.end(), [](int x) { std::cout << x << " "; });
    std::cout << std::endl;
}

int main() {
    int N, C;
    cin >> N >> C;
    vector<vector<int>> D(C, vector<int>(C, 0));
    REP(i, C) {
        REP(j, C) { cin >> D[i][j]; }
    }
    vector<vector<int>> c(N, vector<int>(N, 0));
    REP(i, N) {
        REP(j, N) { cin >> c[i][j]; }
    }
    // cout << "a" << endl;
    // 3とおりのグループに対してそれぞれを
    //あるいろで(C通り)で塗り分けた時のコストを
    //考える
    vector<vector<long long>> cost(3, vector<long long>(C, 0));
    REP(color, C) {
        REP(i, N) {
            REP(j, N) {
                int tmp = i + j;
                tmp %= 3;
                cost[tmp][color] += D[c[i][j] - 1][color];
            }
        }
    }
    long long res = INF;
    for(int i = 0; i < C; i++) {
        for(int j = i + 1; j < C; j++) {
            for(int k = j + 1; k < C; k++) {
                res = min(res, cost[0][i] + cost[1][j] + cost[2][k]);
                res = min(res, cost[1][i] + cost[2][j] + cost[0][k]);
                res = min(res, cost[2][i] + cost[0][j] + cost[1][k]);
            }
        }
    }
    cout << res << endl;
}