Magnus decided to play a classic chess game. Though what he saw in his locker shocked him! His favourite chessboard got broken into 4 pieces, each of size n by n, n is always odd. And what's even worse, some squares were of wrong color. j-th square of the i-th row of k-th piece of the board has color ak, i, j; 1 being black and 0 being white. 

Now Magnus wants to change color of some squares in such a way that he recolors minimum number of squares and obtained pieces form a valid chessboard. Every square has its color different to each of the neightbouring by side squares in a valid board. Its size should be 2n by 2n. You are allowed to move pieces but not allowed to rotate or flip them.

Input

The first line contains odd integer n (1 ≤ n ≤ 100) — the size of all pieces of the board. 

Then 4 segments follow, each describes one piece of the board. Each consists of n lines of n characters; j-th one of i-th line is equal to 1 if the square is black initially and 0 otherwise. Segments are separated by an empty line.

Output

Print one number — minimum number of squares Magnus should recolor to be able to obtain a valid chessboard.

Examples

Input

1
0

0

1

0


Output

1


Input

3
101
010
101

101
000
101

010
101
011

010
101
010


Output

2
#####################################   
import atexit, io, sys, collections, math, heapq
buffer = io.BytesIO() 
sys.stdout = buffer
@atexit.register 
def write(): sys.__stdout__.write(buffer.getvalue())   
##################################### 



n = int(raw_input())
m = []
for i in range(4):
	m.append([map(int,list(raw_input())) for _ in range(n)])
	if i < 3:
		raw_input()
r = [2*n * 2*n]

def dfs(i,m):
	if i == len(m):
		build(m)
		return 
	for j in range(i,len(m)):
		m[i],m[j] = m[j],m[i]
		dfs(i+1, m)
		m[i],m[j] = m[j],m[i]
def cost(t):
	ans = 0
	for i,l in enumerate(t):
		for j in range(len(l)):
			ans +=  l[j] == (i+j) % 2
	return ans	
def build(m):
	a,b = m[:2]
	t = []
	for u,v in zip(a,b):
		t.append(u+v)
	a,b = m[2:]
	for u,v in zip(a,b):
		t.append(u+v)
	r[0] = min(r[0], cost(t))



dfs(0, m)


print r[0]