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 forn(i, x, n) for (int i = x; i < n; i++)
#define all(a) (a).begin(), (a).end()
#define pii pair<int, int>
#define pll pair<ll, ll>
typedef long double ld;
typedef long long ll;

int ans = INT_MAX;
int d[33][33];
int co[500][500];

int calc(vector<int>& v, int i) {
    int cur = 0;
//    cout << i << " =" << endl;
    for (int x = 0; x < v.size(); x++) {
//        cout << v[x] << endl;
        cur += d[v[x] - 1][i];
//        cout << v[x] << " " << d[v[x] - 1][i] << endl;
    }
    return cur;
}

void gen() {exit(0);}
int main() {
    #ifdef DEBUG
    freopen("input", "r", stdin);
//    freopen("output", "w", stdout);
//    gen();
    #endif
    int n, c;
    scanf("%d %d", &n, &c);
    vector<int> x;
    vector<int> y;
    vector<int> z;
    forn(i, 1, c + 1) {
        forn(j, 1, c + 1) {
            scanf(" %d", &d[i - 1][j - 1]);
        }
    }
    forn(i, 1, n + 1) {
        forn(j, 1, n + 1) {
            scanf(" %d", &co[i - 1][j - 1]);
            if (((i + j) % 3) == 0) {
                x.push_back(co[i - 1][j - 1]);
            } else if (((i + j) % 3) == 1) {
                y.push_back(co[i - 1][j - 1]);
            } else {
                z.push_back(co[i - 1][j - 1]);
            }
        }
    }
    int dp[3][33];
    fill(dp[0], dp[0] + (3 * 33), 1e9);
    for (int i = 0; i < c; i++) {
        dp[0][i] = min(dp[0][i], calc(x, i));
        dp[1][i] = min(dp[1][i], calc(y, i));
        dp[2][i] = min(dp[2][i], calc(z, i));
//        cout << dp[0][i] << " " << dp[1][i] << " " << dp[2][i] << endl;
    }
    int ans = INT_MAX;
    forn(i, 0, c) {
        forn(j, 0, c) {
            forn(k, 0, c) {
                if (i != j && j != k && i != k) {
                    int cur = dp[0][i] + dp[1][j] + dp[2][k];
//                    cout << cur << endl;
                    ans = min(ans, cur);
                }
            }
        }
    }
    printf("%d\n", ans);
    return 0;
}