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: Time Limit Exceeded)

#include <bits/stdc++.h>
using namespace std;

template<typename FI>
void parted_rotate(FI first1, FI last1, FI first2, FI last2)
{
 if(first1 == last1 || first2 == last2) return;
 FI next = first2;
 while (first1 != next) {
  std::iter_swap(first1++, next++);
  if(first1 == last1) first1 = first2;
  if (next == last2) {
   next = first2;
  } else if (first1 == first2) {
   first2 = next;
  }
 }
}
 
template<typename BI>
bool next_combination_imp(BI first1, BI last1, BI first2, BI last2)
{
 if(first1 == last1 || first2 == last2) return false;
 auto target = last1; --target;
 auto last_elem = last2; --last_elem;
 // find right-most incrementable element: target
 while(target != first1 && !(*target < *last_elem)) --target;
 if(target == first1 && !(*target < *last_elem)) {
  parted_rotate(first1, last1, first2, last2);
  return false;
 }
 // find the next value to be incremented: *next
 auto next = first2;
 while(!(*target < *next)) ++next;
 std::iter_swap(target++, next++);
 parted_rotate(target, last1, next, last2);
 return true;
}
 
// INVARIANT: is_sorted(first, mid) && is_sorted(mid, last)
template<typename BI>
inline bool next_combination(BI first, BI mid, BI last)
{
 return next_combination_imp(first, mid, mid, last);
}
 
// INVARIANT: is_sorted(first, mid) && is_sorted(mid, last)
template<typename BI>
inline bool prev_combination(BI first, BI mid, BI last)
{
 return next_combination_imp(mid, last, first, mid);
}

int main(void) {
  int N, C;
  cin >> N >> C;
  vector<vector<int>> D(C, vector<int>(C));
  vector<vector<int>> c(N, vector<int>(N));

  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<int> colors;
  for (int i = 0; i < C; i++) {
    colors.push_back(i);
  }

  int ret = 1e9;
  do {
    vector<int> perm(colors.begin(), colors.begin() + 3);
    do {
      int cost = 0;
      // cerr << perm[0] << " " << perm[1] << " " << perm[2] << endl;
      for (int i = 0; i < N; i++) {
        for (int j = 0; j < N; j++) {
          int mod = (i + j) % 3;
          cost += D[c[i][j]][perm[mod]];
        }
      }
      ret = min(ret, cost);
    } while(next_permutation(perm.begin(), perm.end()));
  } while(next_combination(colors.begin(), colors.begin() + 3, colors.end()));
  cout << ret << endl;
  return 0;
}