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: Wrong Answer)

#include <iostream>
#include <string>
#include <vector>
#include <stdio.h>
#include <algorithm>
#define _USE_MATH_DEFINES
#include <cmath>
#include <numeric>
#include <iomanip>
#include <deque>
#include <tuple>
#include <queue>
#include <stack>
#include <map>
#include <set>
#include <unordered_map>
#include <sstream>
#include <boost/multiprecision/cpp_int.hpp>
#include <boost/multiprecision/cpp_dec_float.hpp>

#define rep(i, x) for (int i = 0; i < (int)(x); i++)
#define FOR(i, a, done) for (int i = (a); i < (done); ++i)
#define all(a) (a).begin(), (a).end()
#define F first
#define S second
#define pb push_back
#define pf push_front

namespace mp = boost::multiprecision;
using cpp_int = mp::cpp_int;
using ll = long long;
using namespace std;

ll M = 1000000007;
typedef pair<int, int> P;

int main()
{
    int n, c;
    cin >> n >> c;
    vector<vector<int>> D(c, vector<int>(c));
    rep(i, c) rep(j, c) cin >> D[i][j];
    vector<vector<int>> C(n, vector<int>(n));
    vector<vector<int>> used(3, vector<int>(c));
    rep(i, n)
    {
        rep(j, n)
        {
            cin >> C[i][j];
            C[i][j]--;
            rep(k, 3)
            {
                if ((i + j) % 3 == k)
                {
                    used[k][C[i][j]]++;
                }
            }
        }
    }
    int ans = INT32_MAX;
    //  cout<<ans<<endl;
    rep(i, c)
    {
        int sum = 0;
        rep(ii, c)
        {
            sum += D[ii][i] * used[0][ii];
        }
        rep(j, c)
        {
            if (i != j)
            {
                rep(jj, c)
                {
                    sum += D[jj][j] * used[1][jj];
                }
                rep(k, c)
                {
                    if (k != j && j != i && k != i)
                    {
                        rep(kk, c)
                        {
                            sum += D[kk][k] * used[2][kk];
                        }
                        ans=min(ans,sum);
                    }
                }
            }
        }
    }
    cout<<ans<<endl;
}