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: Wrong Answer)

import numpy as np

def min3(A):
    inf = 10 ** 9
    i0 = np.argmin(A[0])
    i1 = np.argmin(A[1])
    i2 = np.argmin(A[2])
    if abs((i1 - i2) * (i2 - i0) * (i0 - i1)):
        return np.min(A[0]) + np.min(A[1]) + np.min(A[2])
    re0 = re1 = re2 = inf
    if i0 == i1 or i0 == i2:
        mn = np.min(A[0])
        A[0][i0] = inf
        re0 = min3(A)
        A[0][i0] = mn
    if i0 == i1 or i1 == i2:
        mn = np.min(A[1])
        A[1][i1] = inf
        re1 = min3(A)
        A[1][i1] = mn
    if i2 == i1 or i0 == i2:
        mn = np.min(A[0])
        A[2][i2] = inf
        re2 = min3(A)
        A[2][i2] = mn
    return min(re0, re1, re2)

gs, cn = map(int, input().split())
d = [list(map(int, input().split())) for _ in range(cn)]
c = [list(map(int, input().split())) for _ in range(gs)]
cc = [[0] * cn for _ in range(3)]
for i in range(gs):
    ci = c[i]
    for j in range(gs):
        cc[(i + j) % 3][ci[j] - 1] += 1
cc = np.array(cc)
d = np.array(d)
np.dot(cc, d, out=cc)
print(min3(cc))