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<vector>
#include<string>
#include<algorithm>
#include<utility>
#include<cmath>
#include<climits>
#include<queue>
#include<stack>
#include<numeric>
#include<set>
#include<iomanip>
#include<map>
#include<type_traits>
#include<tuple>
#include<deque>
#include<cassert>
#include<bitset>
using namespace std;
typedef long long ll;
typedef pair<int,int> P;
#define rep(i,N) for(ll (i)=0;(i)<(N);(i)++)
#define chmax(x,y) x=max(x,y)
#define chmin(x,y) x=min(x,y)
const int mod = 1000000007;
const int INF = 1001001001;


int main() {
  int n, C;
  cin >> n >> C;
  vector<vector<int>> d(C, vector<int>(C));
  vector<vector<int>> c(n, vector<int>(n));
  rep(i, C) rep(j, C) {
    cin >> d[i][j];
  }
  vector<vector<int>> cnt(3, vector<int>(C));
  rep(i, n) rep(j, n) {
    int t;
    cin >> t;
    --t;
    c[i][j] = t;
    int s = (i + j) % 3;
    cnt[s][t]++;
  }

  int ans = INF;
  rep(i, C)rep(j, C)if(i != j)rep(k, C) {
    if(j == k || k == i) continue;
    int t = 0;
    rep(x, 3) rep(y, C) {
      int next;
      if(x == 0) next = i;
      else if (x == 1) next = j;
      else next = k;
      t += cnt[x][y] * d[y][next];
    }
    ans = min(ans, t);
  }
  cout << ans << endl;
}