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)

/*
  AtCoder Beginner Contest 099
  D - Good Grid
*/

#include<iostream>
#include<vector>
#include<limits>
using namespace std;

int N, C;
int D[30][30];
int c[3][500] = {0};

int main() {
  scanf("%d%d", &N, &C);
  
  for (int i = 0; i < C; ++i) for (int j = 0; j < C; ++j) scanf("%d", &D[i][j]);
  
  int buf;
  for (int i = 0; i < N; ++i) for (int j = 0; j < N; ++j) {
    scanf("%d", &buf);
    for (int k = 0; k < C; ++k) c[((i+1)+(j+1))%3][k] += D[buf-1][k];
  }

  int minval = numeric_limits<int>::max();
  for (int i = 0; i < C; ++i) for (int j = 0; j < C; ++j){
    if(i!=j) {
      for (int k = 0; k < C; ++k){
        if(i!=k && j!=k) {
          int val = c[0][i] + c[1][j] + c[2][k];
          if (val < minval) minval = val;
        }
      }
    }
  }

  printf("%d\n", minval);
  return 0;
}