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::str::FromStr;

pub fn read<T: FromStr>() -> T {
    let stdin = stdin();
    let stdin = stdin.lock();
    let token: String = stdin
        .bytes()
        .map(|c| c.expect("failed to read char") as char)
        .skip_while(|c| c.is_whitespace())
        .take_while(|c| !c.is_whitespace())
        .collect();
    token.parse().ok().expect("failed to parse token")
}

use std::cmp::min;

fn main() {
    let n: usize = read();
    let c: usize = read();
    let mut transform: Vec<Vec<i64>> = vec![vec![0; c]; c];

    for i in 0..c {
        for j in 0..c {
            transform[i][j] = read();
        }
    }

    let mut map: Vec<Vec<i64>> = vec![vec![0; n]; n];
    for i in 0..n {
        for j in 0..n {
            map[i][j] = read();
        }
    }
    // for k in 0..c {
    //     for i in 0..c {
    //         for j in 0..c {
    //             transform[i][j] = min(transform[i][j], transform[i][k] + transform[k][j]);
    //         }
    //     }
    // }

    let mut memo: Vec<Vec<i64>> = vec![vec![0; 3]; c];

    for i in 0..n {
        for j in 0..n {
            for k in 0..c {
                let oc = map[i][j];
                let ncc = transform[oc as usize - 1][k];
                memo[k][(i + j) % 3] += ncc;
            }
        }
    }
    // for i in 0..c {
    //     println!("{:?}", memo[i]);
    // }

    let mut minv = std::i64::MAX;
    for i in 0..c {
        for j in 0..c {
            for k in 0..c {
                if i == j || j == k || k == i {
                    continue;
                }
                minv = min(minv, memo[i][0] + memo[j][1] + memo[k][2]);
                minv = min(minv, memo[i][0] + memo[j][2] + memo[k][1]);
                minv = min(minv, memo[i][1] + memo[j][2] + memo[k][0]);
                minv = min(minv, memo[i][1] + memo[j][0] + memo[k][2]);
                minv = min(minv, memo[i][2] + memo[j][1] + memo[k][0]);
                minv = min(minv, memo[i][2] + memo[j][0] + memo[k][1]);
            }
        }
    }
    println!("{}", minv);
}