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::vec::Vec;
use std::cmp::*;

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

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

fn main() {
  let (N, C) = {
    let v: Vec<usize> = read_vec();
    (v[0], v[1])
  };

  let mut D: Vec<Vec<i64>> = Vec::new();
  for _ in 0..C {
    let d: Vec<i64> = read_vec();
    D.push(d);
  }

  let mut c: Vec<Vec<i64>> = Vec::new();
  for _ in 0..N {
    let v: Vec<i64> = read_vec::<i64>().iter().map(|a| a - 1).collect();
    c.push(v);
  }

  // 前計算
  let mut sum: Vec<Vec<i64>> = vec![vec![0; C]; 3];
  for i in 0..C {
    for j in 0..N {
      for k in 0..N {
        if (j + k) % 3 == 0 {
          sum[0][i] += D[c[j][k] as usize][i];
        }
        else if (j + k) % 3 == 1 {
          sum[1][i] += D[c[j][k] as usize][i];
        }
        else {
          sum[2][i] += D[c[j][k] as usize][i];
        }
      }
    }
  }

  // 全探索
  let mut res = 1_000_000_000;
  for i in 0..C {
    for j in 0..C {
      if i == j {
        continue;
      }
      for k in 0..C {
        if j == k || i == k{
          continue;
        }

        res = min(sum[0][i] + sum[1][j] + sum[2][k], res);

      }
    }
  }
  println!("{}", res);
}