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 <vector>
#include <algorithm>
#include <ctime>
#include <cmath>
#include <string>
#include <stdlib.h>
#include <fstream>
#include <set>
#include <map>
#include <queue>
#include <stack>
#include <deque>
#include <iterator>
#include <iomanip>
#include <climits>
using namespace std;

int main(){
    long long n, c;
    cin >> n >> c;
    vector<vector<long long> > d(c, vector<long long>(c)), a(n, vector<long long>(n));
    for(int i = 0; i < c; ++i)
        for(int j = 0; j < c; ++j) cin >> d[i][j];

    vector<map<long long, long long> > cnt(3);

    for(int i = 0; i < n; ++i)
        for(int j = 0; j < n; ++j){
            cin >> a[i][j];
            cnt[(i + j + 2) % 3][a[i][j]]++;
        }
    long long ans = 0;
    vector<vector<pair<long long, long long> > > b(3, vector<pair<long long, long long> >(c));
    for(int j = 0; j < 3; ++j){
        for(int i = 0; i < c; ++i){
            long long sum = 0;
            for(auto x: cnt[j]){
                sum += x.second * d[x.first - 1][i];
            }
            b[j][i] = {sum, i};
        } 
    }
    for(int i = 0; i < 3; ++i){
        sort(b[i].begin(), b[i].end());
    }
    vector<int> ind = {0, 0, 0};
    while(b[0][ind[0]].second == b[1][ind[1]].second || b[1][ind[1]].second == b[2][ind[2]].second || b[0][ind[0]].second == b[2][ind[2]].second){
        for(int i = 0; i < 3; ++i){
            for(int j = 0; j < 3; ++j){
                if(i == j) continue;
                if(b[i][ind[i]].second == b[j][ind[j]].second){
                    if(b[2][ind[i]].first > b[j][ind[j]].first) ind[j]++;
                    else ind[i]++;
                }
            }
        }
    }
    cout << b[0][ind[0]].first + b[1][ind[1]].first + b[2][ind[2]].first;
}