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)

#define _USE_MATH_DEFINES
#define _CRT_SECURE_NO_WARNINGS
#include <bits/stdc++.h>
using namespace std;
//
#define int long long
#define ALL(x) (x).begin(), (x).end()
#define UNIQUE(v) (v).erase(unique((v).begin(), (v).end()), (v).end());
#define m0(x) memset((x), 0, sizeof(x))
#define mm(x) memset((x), -1, sizeof(x))
// bit_macro
#define bit(n) (1LL << (n))
#define bitset(a, b) (a) |= (1 << (b))
#define bitunset(a, b) (a) |= ~(1 << (b))
#define bitcheck(a, b) ((a) >> (b)) & 1
#define bitcount(a) __builtin_popcountll((a))
//
typedef long long lint;
typedef pair<int, int> P;
typedef tuple<int, int, int> TP;
typedef vector<int> vec;
typedef vector<vec> mat;
//
const int INF = (int)1e18;
const int MOD = (int)1e9 + 7;
const double EPS = (double)1e-10;
const int dx[] = {-1, 0, 0, 1};
const int dy[] = {0, -1, 1, 0};
//
template <typename T> void chmax(T &a, T b) { a = max(a, b); }
template <typename T> void chmin(T &a, T b) { a = min(a, b); }
//
struct Accelerate_Cin {
  Accelerate_Cin() {
    cin.tie(0);
    ios::sync_with_stdio(0);
    cout << fixed << setprecision(20);
  };
};

int N, C;
int D[1010][1010], c[1010][1010];
int sum[5][35]; // sum[i][c]:=
                // (x+i)%3==iを満たすマスを,全てcに変更したときの総コスト

signed main() {
  cin >> N >> C;
  for (int i = 1; i <= C; i++) {
    for (int j = 1; j <= C; j++) {
      cin >> D[i][j];
    }
  }
  for (int i = 1; i <= N; i++) {
    for (int j = 1; j <= N; j++) {
      cin >> c[i][j];
    }
  }

  m0(sum);

  for (int co = 1; co <= C; co++) {

    for (int i = 1; i <= N; i++) {
      for (int j = 1; j <= N; j++) {
        int temp = (i + j) % 3;
        sum[temp][co] += D[c[i][j]][co];
      }
    }
  }

  int ans = INF;
  for (int i = 1; i <= C; i++) {
    for (int j = 1; j <= C; j++) {
      for (int k = 1; k <= C; k++) {
        if (i == j || j == k || k == i) {
          continue;
        }

        int temp = sum[0][i] + sum[1][j] + sum[2][k];
        chmin(ans, temp);
      }
    }
  }

  cout << ans << endl;
  return 0;
}