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 <vector>
#include <iostream>
#include <algorithm>
#include <string>
#include <set>
#include <vector>
#include <queue>
using namespace std;

int main(){
	int n, c;
	cin >> n >> c;
	vector<vector<int>> d = vector<vector<int>>(c, vector<int>(c, 0));
	for (int i = 0; i < c; ++i) {
		for (int j = 0; j < c; ++j) {
			int dij;
			cin >> dij;
			d[i][j] = dij;
		}
	}
	vector<vector<int>> cmap = vector<vector<int>>(n, vector<int>(n, 0));
	for (int i = 0; i < n; ++i) {
		for (int j = 0; j < n; ++j) {
			int cij;
			cin >> cij;
			cij--;
			cmap[i][j] = cij;
		}
	}
	vector<vector<int>> result = vector<vector<int>>(3, vector<int>(c,0));
	for(int ct =0;ct<c;++ct){
		for (int i = 0; i < n; ++i) {
			for (int j = 0; j < n; ++j) {
				int line = (i + j) % 3;
				int cc = cmap[i][j];
				if(cc!=ct)
					result[line][ct] += d[cc][ct];
			}
		}
	}
	int best = 500*500*1000+1;
	for (int call = 0; call < c*c*c; call++) {
		int c1 = call % c;
		int c2 = (call / c) % c;
		int c3 = (call / c / c) % c;
		if (c1 == c2 || c2 == c3 || c3 == c1) continue;
		best = min(best, result[0][c1] + result[1][c2] + result[2][c3]);
	}
	cout << best << endl;
	return 0;
}