Iahub is very proud of his recent discovery, propagating trees. Right now, he invented a new tree, called xor-tree. After this new revolutionary discovery, he invented a game for kids which uses xor-trees.

The game is played on a tree having *n* nodes, numbered from 1 to *n*. Each node *i* has an initial value *init*<sub class="lower-index">*i*</sub>, which is either 0 or 1. The root of the tree is node 1.

One can perform several (possibly, zero) operations on the tree during the game. The only available type of operation is to pick a node *x*. Right after someone has picked node *x*, the value of node *x* flips, the values of sons of *x* remain the same, the values of sons of sons of *x* flips, the values of sons of sons of sons of *x* remain the same and so on.

The goal of the game is to get each node *i* to have value *goal*<sub class="lower-index">*i*</sub>, which can also be only 0 or 1. You need to reach the goal of the game by using minimum number of operations.
The first line contains an integer *n* (1 ≤ *n* ≤ 10<sup class="upper-index">5</sup>). Each of the next *n* - 1 lines contains two integers *u*<sub class="lower-index">*i*</sub> and *v*<sub class="lower-index">*i*</sub> (1 ≤ *u*<sub class="lower-index">*i*</sub>, *v*<sub class="lower-index">*i*</sub> ≤ *n*; *u*<sub class="lower-index">*i*</sub> ≠ *v*<sub class="lower-index">*i*</sub>) meaning there is an edge between nodes *u*<sub class="lower-index">*i*</sub> and *v*<sub class="lower-index">*i*</sub>. 

The next line contains *n* integer numbers, the *i*-th of them corresponds to *init*<sub class="lower-index">*i*</sub> (*init*<sub class="lower-index">*i*</sub> is either 0 or 1). The following line also contains *n* integer numbers, the *i*-th number corresponds to *goal*<sub class="lower-index">*i*</sub> (*goal*<sub class="lower-index">*i*</sub> is either 0 or 1).
# python3

mapped_dict = {}

for x in range(1, int(input().strip())):
	a, b = [int(x) for x in input().split()]
	if a in mapped_dict:
		mapped_dict[a].append(b)
	else:
		mapped_dict[a] = []
		mapped_dict[a].append(b)

def flip_num(x):
	if x == 1:
		return 0
	else: 
		return 1

def flip_evens(nodes, initial):
	queue = []
	queue.append(initial)
	while len(queue) != 0:
		item = queue.pop()
		nodes[item] = flip_num(nodes[item])
		dne = False
		if item in mapped_dict:
			for next_node in mapped_dict[item]:
				if next_node in mapped_dict:
					for second_node in mapped_dict[next_node]:
						queue.append(second_node)

from itertools import permutations

def allPermutations(seq):
    return (x for i in range(len(seq),0,-1) for x in permutations(seq, i))

import copy
def bfs(nodes, goal):
	values = range(len(nodes))
	all_permutations = allPermutations(values)
	found = False
	for permut in all_permutations:
		if found: break
		cur_nodes = copy.copy(nodes)
		for num in permut:
			if found: break
			flip_evens(cur_nodes, num)
			if cur_nodes == goal:
				found = True
				return permut
	return []

init = [int(x) for x in input().split()]
goal = [int(x) for x in input().split()]

solution = bfs(init, goal)
for sol in solution:
	print(sol+1, end=" ")