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: Runtime Error)

#include<bits/stdc++.h>
#define FOR(i, a, b) for (int i = a; i < b; ++i)
#define rep(N) for (int i = 0; i < N; ++i)
#define Rep(a, b) for (int i = a; i < b; ++i)
#define For(i, N) for (int i = 0; i < N; ++i)
#define all(v) v.begin(), v.end()
#define rev(v) v.rbegin(), v.rend()
#define makei(N) int N; cin >> N;
#define makel(N) ll N; cin >> N;
#define makes(s) string s; cin >> s;
#define maked(d) double d; cin >> d;
#define makev(v, N) vi v(N); rep(N)cin >> v[i];
#define mod 1000000007
#define re return 0
#define distance(x1,y1,x2,y2) sqrt((x1 - x2)*(x1 - x2) + (y1 - y2)*(y1 - y2));

using namespace std;
using ll = long long int;
using vi = vector<int>;
using vvi = vector<vector<int>>;
using vll = vector<ll>;
using vb = vector<bool>;
using vvb = vector<vector<bool>>;
using vs = vector<string>;
using pii = pair<int, int>;
using pis = pair<int, string>;
using msi = map<string, int>;

template<typename T>
void say(T s) {
	cout << s << "\n";
}

template<typename T>
void say(vector<T> s) {
	auto itr = s.begin();
	cout << *(itr++);
	while (itr != s.end()) {
		cout << " " << *(itr++);
	}
	cout << "\n";
}

void yn(bool b) {
	if (b)say("Yes");
	else say("No");
}

template<typename T>
void maxi(T &a, T b) {
	a = max(a, b);
}

template<typename T>
void mini(T &a, T b) {
	a = min(a, b);
}

int main() {
	makei(N);
	makei(C);
	vvi d(C);
	vvi ans;
	vvi c(3);
	For(i, 3)For(j, C)ans[i].push_back(0);
	For(i, C) {
		For(j, C) {
			makei(D);
			d[i].push_back(D);
		}
	}
	For(i, N) {
		For(j, N) {
			makei(D);
			c[(i + j) % 3].push_back(D - 1);
		}
	}
	For(i, 3) {
		For(j, C) {
			for (int k : c[i]) {
				ans[i][j] += d[k][j];
			}
		}
	}
	int Ans = INT32_MAX;
	For(i, C) {
		For(j, C) {
			For(k, C) {
				if (i == j || j == k || k == i)continue;
				mini(Ans, ans[0][i] + ans[1][j] + ans[2][k]);
			}
		}
	}
	say(Ans);
}