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)

#[allow(dead_code)]
fn getline() -> String {
    let mut res = String::new();
    std::io::stdin().read_line(&mut res).ok();
    res
}

#[allow(unused_macros)]
macro_rules! readl {
    ($t: ty) => {
        {
            let s = getline();
            s.trim().parse::<$t>().unwrap()
        }
    };
    ($( $t: ty),+ ) => {
        {
            let s = getline();
            let mut iter = s.trim().split(' ');
            ($(iter.next().unwrap().parse::<$t>().unwrap(),)*)
        }
    };
}

#[allow(unused_macros)]
macro_rules! readlvec {
    ($t: ty) => {
        {
            let s = getline();
            let iter = s.trim().split(' ');
            iter.map(|x| x.parse().unwrap()).collect::<Vec<$t>>()
        }
    }
}

#[allow(unused_macros)]
macro_rules! debug { ($x: expr) => { println!("{}: {:?}", stringify!($x), $x) } }
// macro_rules! debug { ($x: expr) => () }

#[allow(dead_code)]
fn show<T>(iter: T) -> String
where
    T: Iterator,
    T::Item: std::fmt::Display
{
    let mut res = iter.fold(String::new(), |sum, e| sum + &format!("{} ", e));
    res.pop();
    res
}

#[allow(unused_imports)]
use std::cmp::{max, min};
#[allow(unused_imports)]
use std::collections::{BTreeMap, BTreeSet, BinaryHeap, HashMap, HashSet, VecDeque};
#[allow(unused_imports)]
use std::collections::btree_map::Entry::{Occupied, Vacant};

fn main() {
    use std::io::Write;
    let out = std::io::stdout();
    let mut out = std::io::BufWriter::new(out.lock());
    macro_rules! printf { ($($arg:tt)*) => (write!(out, $($arg)*).unwrap()); }
    macro_rules! printfln { () => (writeln!(out).unwrap()); ($($arg:tt)*) => (writeln!(out, $($arg)*).unwrap()); }
    
    let (n, c) = readl!(usize, usize);
    let d: Vec<_> = (0..c).map(|_| readlvec!(i64)).collect();
    
    let cv: Vec<Vec<_>> = (0..n).map(|_| readlvec!(i64).into_iter().map(|x| x-1).collect()).collect();
    


    let mut cov = vec![vec![0; c]; 3];
    for (i, row) in cv.iter().enumerate() {
        for (j, &e) in row.iter().enumerate() {
            cov[(i+j)%3][e as usize] += 1;
        }
    }
    let cov = cov;

    let inf = 1_000_000_000_000;
    let mut ans = inf;
    for c0 in 0..c {
        for c1 in 0..c { 
            for c2 in 0..c {
                if c0 == c1 || c1 == c2 || c0 == c2 {
                    continue;
                }
                let ccv = [c0, c1, c2];
                let mut cost = 0;
                for (i, cr) in cov.iter().enumerate() {
                    for (&ce, dr) in cr.iter().zip(d.iter()) {
                        cost += ce*dr[ccv[i]];
                    }
                }
                ans = min(ans, cost);
            }
        }
    }
    printfln!("{}", ans);
}