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;

using ll = long long int;

const int MAX = (int)(1e5 + 5);
const ll INF = (ll)(1e10 + 5);

const int MAX_N = (int)(5e2 + 5);
const int MAX_C = (int)(3e1 + 5);

int n, c;
int diff[MAX_C][MAX_C];
int grid[MAX_N][MAX_N];

int cnt[3][MAX_C];

ll ans;

int main(void) {
  // Here your code !
  scanf("%d %d", &n, &c);

  for (int i = 1; i <= c; ++i) {
    for (int j = 1; j <= c; ++j) {
      scanf("%d", &(diff[i][j]));
    }
  }

  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j) {
      scanf("%d", &(grid[i][j]));
    }
  }

  for (int i = 1; i <= n; ++i) {
    for (int j = 1; j <= n; ++j) {
      cnt[(i + j) % 3][grid[i][j]] += 1;
    }
  }

  ans = INF;

  for (int i = c; i >= 1; --i) {
    for (int j = c; j >= 1; --j) {
      if (i == j) continue;
      for (int k = c; k >= 1; --k) {
        if (i == k || j == k) continue;

        ll tmp = 0LL;

        for (int col = 1; col <= c; ++col) {
          tmp += cnt[0][col] * diff[col][i];
          tmp += cnt[1][col] * diff[col][j];
          tmp += cnt[2][col] * diff[col][k];
        }

        ans = min(ans, tmp);
      }
    }
  }

  printf("%lld\n", ans);

  return 0;
}