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: Time Limit Exceeded)

#include <bits/stdc++.h>
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>

using namespace std;

typedef long long int lli;

template<class T>bool chmax(T &a, const T &b) { if( a < b ) { a = b; return 1; } return 0;}
template<class T>bool chmin(T &a, const T &b) { if( b < a ) { a = b; return 1; } return 0;}

#ifdef LOCAL
	#define eprintf(...) printf(__VA_ARGS__)
#else 
	#define eprintf(...) 1
#endif

#define ROUNDUP(a, b) 		( ((a) + ((b) - 1) ) / b)
#define SET_ZERO(a)	 		( memset((a), 0, sizeof(a)))
#define SET_NUM_ARRAY(a,b)	( memset((a), b, sizeof(a)))
//#define int long long int //dont forget to change int to signed


int main() {
	int N, C;
	scanf("%d %d", &N, &C);
	int D[30][30];
	for (int i = 0; i < C; i++) {
		for (int j = 0; j < C; j++) {
			scanf("%d", &D[i][j]);
		}
	}
	vector<int> v0;
	vector<int> v1;
	vector<int> v2;
	for (int i = 0; i < N; i++) {
		for (int j = 0; j < N; j++) {
			int c;
			scanf("%d", &c);
			int t = (i-1+j-1)%3;
			if (t == 0) v0.push_back(c);
			else if (t == 1) v1.push_back(c);
			if (t == 2) v2.push_back(c);
		}
	}

	lli ans = INT_MAX;
	for (int i = 0; i < C; i++) {
		for (int j = 0; j < C; j++) {
			for (int k = 0; k < C; k++) {
				if (i == j || j == k || k == i) continue;
				lli sum = 0;
				vector<int>::iterator it = v0.begin();
				while (it != v0.end()) {
					int t = *it;
					sum += D[t][C];	
				}
				it = v1.begin();
				while (it != v1.end()) {
					int t = *it;
					sum += D[t][C];	
				}
				it = v2.begin();
				while (it != v2.end()) {
					int t = *it;
					sum += D[t][C];	
				}
				chmin(ans, sum);
			}
		}
	}
	printf("%lld\n", ans);
	return 0;
}