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 Python (3.4.3) (judge verdict: Accepted)

#!/usr/bin/env python3
import sys, math, fractions, itertools


def solve(N: int, C: int, D: "List[List[int]]", c: "List[List[int]]"):
    """
    (x+y)%3が等しければ色も等しいので良いグリッドには色は3色しかない。
    これを利用して30C3種類全探索。
    """
    t = [[0 for _ in range(30)] for _ in range(3)]
    for i in range(N):
        for j in range(N):
            t[(i+j)%3][c[i][j]-1] += 1

    res = 1 << 30
    for i in range(C):
        for j in range(C):
            if i != j:
                for k in range(C):
                    if i != k and j != k:
                        tt = 0
                        for l in range(C):
                            tt += D[l][i] * t[0][l]
                        for l in range(C):
                            tt += D[l][j] * t[1][l]
                        for l in range(C):
                            tt += D[l][k] * t[2][l]
                        if tt < res:
                            res = tt
    print(res)
    return


# Generated by 1.1.4 https://github.com/kyuridenamida/atcoder-tools  (tips: You use the default template now. You can remove this line by using your custom template)
def main():
    def iterate_tokens():
        for line in sys.stdin:
            for word in line.split():
                yield word
    tokens = iterate_tokens()
    N = int(next(tokens))  # type: int
    C = int(next(tokens))  # type: int
    D = [ [ int(next(tokens)) for _ in range(C) ] for _ in range(C) ]  # type: "List[List[int]]"
    c = [ [ int(next(tokens)) for _ in range(N) ] for _ in range(N) ]  # type: "List[List[int]]"
    solve(N, C, D, c)

if __name__ == '__main__':
    main()