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 (Clang 3.8.0) (judge verdict: Accepted)

#include <iostream>
#include <iomanip>
#include <string>
#include <sstream>
#include <fstream>
#include <cassert>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <numeric>
#include <utility>
#include <vector>
#include <queue>
#include <map>
#include <set>
#include <unordered_map>
#include <unordered_set>
#include <bitset>
#include <complex>

using namespace std;

using ll = long long;
using ld = long double;
using pii = pair<int, int>;

constexpr int MAXN = 505;
int n, c;
int d[MAXN][MAXN], a[MAXN][MAXN];

int freq[3][MAXN];

ll score(int idx, int dest) {
    ll res = 0LL;
    for (int cc = 1; cc <= c; ++cc) {
        res += 1LL * freq[idx][cc] * d[cc][dest];
    }

    return res;
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0); cout.tie(0);

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

    for (int i = 0; i < n; ++i) {
        for (int j = 0; j < n; ++j) {
            scanf(" %d", &a[i][j]);
            ++freq[(i + j) % 3][a[i][j]];
        }
    }

    ll best = 1e18;

    for (int c1 = 1; c1 <= c; ++c1) {
        ll cur_first = score(0, c1);
        if (cur_first > best) continue;
        for (int c2 = 1; c2 <= c; ++c2) {
            if (c1 == c2) continue;

            ll cur_second = cur_first + score(1, c2);
            if (cur_second > best) continue;

            for (int c3 = 1; c3 <= c; ++c3) {
                if (c1 == c3 or c2 == c3) continue;
                best = min(best, cur_second + score(2, c3));
            }
        }
    }

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

    return 0;
}