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;

#define  range(i, l, r) for (int i = (int)(l); i < (int)(r); ++(i))
#define rrange(i, l, r) for (int i = (int)(r) - 1; i >= (int)(l); --(i))

#define  whole(f, x, ...) ([&](decltype((x)) container) { return (f)(  begin(container),  end(container), ## __VA_ARGS__); })(x)
#define rwhole(f, x, ...) ([&](decltype((x)) container) { return (f)( rbegin(container), rend(container), ## __VA_ARGS__); })(x)

#define debug(x) cerr << "(" << __LINE__ << ")" << #x << ": " << (x) << endl

using int32 = int;
using int64 = long long;

const int   mod   = 1e9 + 7;
const int32 inf   = 1001001001;
const int64 infll = 1001001001001001001ll;

const int dx[] = { 0, -1,  1,  0, -1,  1, -1,  1}; 
const int dy[] = {-1,  0,  0,  1, -1, -1,  1,  1};

struct IoSetup { IoSetup(int x = 10){ cin.tie(0); ios::sync_with_stdio(0); cout << fixed << setprecision(x); cerr << fixed << setprecision(x); } } iosetup;

template <typename T = int64> T input() { T x; cin >> x; return x; }

template <typename T> ostream &operator<<(ostream &os, vector<T> &v) { range(i, 0, v.size()) { os << v[i] << (i + 1 != v.size() ? " " : ""); } return os; } 
template <typename T> istream &operator>>(istream &is, vector<T> &v) { for (T &in : v) is >> in; return is; }
template <typename T1, typename T2> ostream &operator<<(ostream &os, pair<T1, T2> p) { os << p.first << " " << p.second; return os; }
template <typename T1, typename T2> istream &operator>>(istream &is, pair<T1, T2> &p) { is >> p.first >> p.second; return is; }

template <typename T> vector<T> make_v(size_t a, T b) { return vector<T>(a, b); }
template <typename... Ts> auto make_v(size_t a, Ts... ts) { return vector<decltype(make_v(ts...))>(a, make_v(ts...)); }

template <typename T1, typename T2> inline bool chmax(T1 &a, T2 b) { return a < b && (a = b, true); }
template <typename T1, typename T2> inline bool chmin(T1 &a, T2 b) { return a > b && (a = b, true); }

void solver() {
  int N = input(), C = input();
  auto D = make_v(C, C, 0);
  cin >> D;

  auto c = make_v(N, N, 0);
  cin >> c;
  for (auto &s : c) for (auto &t : s) t--;

  auto v = make_v(3, C, 0);
  range(i, 0, N) range(j, 0, N) {
    v[(i + j) % 3][c[i][j]]++;
  }

  int ans = inf;
  range(i, 0, C) range(j, 0, C) range(k, 0, C) {
    if (i == j || j == k || k == i) continue;

    int sum = 0;
    range(x, 0, C) {
      sum += D[x][i] * v[0][x] +
             D[x][j] * v[1][x] + 
             D[x][k] * v[2][x];
    }

    chmin(ans, sum);
  }

  cout << ans << endl;
}

signed main(int argc, char *argv[]) {
  solver();
}