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 <algorithm>
using namespace std;
using int64 = int64_t;
#define rep(i, n) for (int i = 0; i < (int)(n); ++i)
#define print(x) cout << (x) << '\n'

int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(nullptr);
    
    int n, c; cin >> n >> c;
    int64 D[40][40] = {};
    int64 C[510][510] = {};

    rep(i, c){
        rep(j, c) cin >> D[i][j];
    }

    rep(i, n){
        rep(j, n) cin >> C[i][j];
    }

    int sum0[40] = {}, sum1[40] = {}, sum2[40] = {};

    for (int i = 1; i <= c; ++i){
        for (int j = 0; j < n; ++j){
            for (int k = 0; k < n; ++k){
                if ((j + k) % 3 == 0){
                    sum0[i] += D[C[j][k] - 1][i - 1];
                }
                else if ((j + k) % 3 == 1){
                    sum1[i] += D[C[j][k] - 1][i - 1];
                }
                else if ((j + k) % 3 == 2){
                    sum2[i] += D[C[j][k] - 1][i - 1];
                }
            }
        }
    }

    int ans = 1e9;

    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){
                    ans = min(ans, sum0[i] + sum1[j] + sum2[k]);
                }
            }
        }
    }

    print(ans);
    
    return 0;
}