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 <bits/stdc++.h>
using namespace std;
const int INF = (1 << 30) - 1;
int main()
{
    int N, C;
    int D[33][33];
    int c[505][505];

    cin >> N >> C;
    for ( int i = 0; i < C; i++ ) {
        for ( int j = 0; j < C; j++ ) {
            cin >> D[i][j];
        }
    }

    for ( int i = 0; i < N; i++ ) {
        for ( int j = 0; j < N; j++ ) {
            cin >> c[i][j]; --c[i][j];
        }
    }
    vector < pair < int , int >  > iwa[3];
    for ( int g = 0; g < 3; g++ ) {
        for ( int p = 0; p < C; p++ ) {
            int sum = 0;
            for ( int i = 0; i < N; i++ ) {
                for ( int j = (g - i + 999999999) % 3; j < N; j += 3 ) {
                    sum += D[c[i][j]][p];
                }
            }
            iwa[g].emplace_back(sum, p);
        }
        sort(begin(iwa[g]), end(iwa[g]));
    }

    vector < int > p = { 0, 1, 2 };
    int minv = INF;
    do {
        int sum = 0;
        bool flag[33] = {};
        for ( int pos : p ) {
            for ( auto &x : iwa[pos] ) {
                if ( !flag[x.second] ) {
                    sum += x.first;
                    flag[x.second] = true;
                    break;
                }
            }
        }
        minv = min(minv, sum);
    } while ( next_permutation(begin(p), end(p)) );

    cout << minv << endl;
    return ( 0 );
}