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
buf = input().split()
N = int(buf[0])
C = int(buf[1])
D = np.zeros((C, C))
for i in range(C):
	buf = input().split()
	for j in range(C):
		D[i][j] = int(buf[j])
		
c = np.zeros(3)
for i in range(2, N+2):
	c[(i%3)] += i-1
for i in range(N+2, 2*N+1):
	c[(i%3)] += N-1-(i-(N+2))
	
grid = -np.ones((3, int(np.max(c))))
groopcount = np.zeros(3)
for i in range(N):
	buf = input().split()
	for j in range(N):
		groop = (i+1+j+1)%3
		grid[groop][int(groopcount[groop])] = int(buf[j])-1
		groopcount[groop] += 1
#print(grid)
iwakan = np.zeros((3, C))

for groop in range(3):
	for color in range(C):
		#print(c3 + ((N*N)%3)//(groop+1))
		for i in range(int(c[groop])):
			#print(groop, color, i)
			#print(grid[groop][i])
			#print(D[int(grid[groop][i]-1)][color])
			iwakan[groop][color] += D[int(grid[groop][i])][color]
#print(iwakan)

miniwakan = np.zeros((3, 3))
cminiwakan = np.zeros((3, 3))
for groop in range(3):
	buf = iwakan[groop]
	for best in range(3):
		miniwakan[groop][best] = np.min(buf)
		mcolor = np.argmin(buf)
		cminiwakan[groop][best] = mcolor
		buf[mcolor] = 1000*c[groop]
#print(miniwakan)
#print(cminiwakan)
min = 1000*np.max(c)
for i in range(3):
	for j in range(3):
		for k in range(3):
			bufmin = miniwakan[0][i] + miniwakan[1][j] + miniwakan[2][k]
			c0 = cminiwakan[0][i]
			c1 = cminiwakan[1][j]
			c2 = cminiwakan[2][k]
			if((c0!=c1) and (c0!=c2) and (c1!=c2) and bufmin<min):
				min = bufmin

print(int(min))
					
			

# your code goes here