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 Rust (1.15.1) (judge verdict: Accepted)

use std::io::*;
use std::cmp;

fn read<T: std::str::FromStr>() -> T {
    let stdin = stdin();
    let mut buf = String::new();
	stdin.lock().read_line(&mut buf);
	buf.trim().parse().ok().unwrap()
}

fn read_vec<T: std::str::FromStr>() -> Vec<T> {
	read::<String>().trim().split_whitespace()
        .map(|w| w.parse().ok().unwrap()).collect()
}

fn read_mat<T: std::str::FromStr>(n: usize) -> Vec<Vec<T>> {
    (0..n).map(|_| read_vec()).collect()
}

fn main() {
    let v = read_vec::<usize>();
    let (n,m) = (v[0],v[1]);
    let d = read_mat::<usize>(m);
    let c = read_mat::<usize>(n);
    let mut min3: Vec<Vec<(usize,usize)>> = vec![vec![(usize::max_value(),0);3];3];
    for col in 0..m {
        let mut tv: Vec<usize> = vec![0;3];
        for i in 0..n {
            for j in 0..n {
                tv[(i+j)%3] += d[c[i][j]-1][col];
            }
        }
        min3[0].push((tv[0],col));
        min3[0].sort_by(|a,b| a.0.cmp(&b.0));
        min3[0].remove(3);
        min3[1].push((tv[1],col));
        min3[1].sort_by(|a,b| a.0.cmp(&b.0));
        min3[1].remove(3);
        min3[2].push((tv[2],col));
        min3[2].sort_by(|a,b| a.0.cmp(&b.0));
        min3[2].remove(3);
    }
    let mut ans = usize::max_value();
    for i in 0..3 {
        for j in 0..3 {
            for k in 0..3 {
                if min3[0][i].1 != min3[1][j].1 && min3[1][j].1 != min3[2][k].1 && min3[0][i].1 != min3[2][k].1 {
                    ans = cmp::min(ans,min3[0][i].0 + min3[1][j].0 + min3[2][k].0);
                }
            }
        }
    }
    println!("{}",ans);
}