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 <iostream>
#include <string>
#include <algorithm>
#include <vector>
#include <array>
#include <queue>
#include <map>
#include <set>
#include <sstream>
#include <stdio.h>
#include <cstdio>
#include <cstring>

using namespace std;

#define Rep(i,a,b) for(int i = a; i < b; i++)
#define rep(i,b) Rep(i,0,b)
#define rrep(i,a) for(int i = a; i >= 0; i--)
#define allof(a) (a).begin(), (a).end()

typedef long long ll;
const int inf = 1e9 + 7;
const ll infll = 1ll << 60ll;
const ll mod = 1e9 + 7;
// 0~3までは右左下上 4~7までは斜め
constexpr int dx[] = { 1, 0, -1, 0, 1, 1, -1, -1 };
constexpr int dy[] = { 0, -1, 0, 1, 1, -1, -1, 1 };

int main() {
	cin.tie(0);
	ios::sync_with_stdio(false);

	int n, c;
	cin >> n >> c;
	vector<vector<int>> d(c, vector<int>(c));
	vector<vector<int>> co(n, vector<int>(n));

	rep(i, c) {
		rep(j, c) {
			cin >> d[i][j];
		}
	}
	rep(i, n) {
		rep(j, n) {
			cin >> co[i][j];
		}
	}

	// 必要な色の種類は3種類のみ(1マス以外)
	// どの一色使うか決める
	int ans = inf;

	vector<vector<int>>cNum(3, vector<int>(c, 0));

	rep(i, n) {
		rep(j, n) {
			rep(k, 3) {
				if ((i + j + 2) % 3 == k) {
					cNum[k][co[i][j] - 1]++;
				}
			}
		}
	}

	rep(i, c) {
		rep(j, c) {
			if (j == i) continue;
			rep(k, c) {
				if (k == i || k == j) continue;
				// 0番目にi色を
				// 1番目にj色を
				// 2番目にk色を使う
				int temp = 0;
				rep(l, c) {
					temp += d[l][i] * cNum[0][l];
					temp += d[l][j] * cNum[1][l];
					temp += d[l][k] * cNum[2][l];
				}
				ans = min(ans, temp);
			}
		}
	}

	cout << ans << endl;

	return 0;
}