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).
from collections import defaultdict
n = int(input())

d = defaultdict(list)
def DFSop(d1,x,goal,state,c,visited,l):

	if c == 1:
		state[x-1] = 1
	elif c == 0:
		state[x-1] = 1
	if goal[x-1] != state[x-1]:
		count = 1
		c = goal[x-1]
		l.append(x)
	else:
		count = 0
	visited[x] = 1
	for i in d1[x]:

		if visited[i] == 0:
			DFSop(d1,i,goal,state,c,visited,l)
	return l

def DFS(d,x,parent,visited):

	stack = [(x,0)]
	d1 = defaultdict(list)

	while len(stack):

		temp,parent = stack.pop()
		visited[temp] = 1
		# print(temp,parent)
		for i in d[temp]:

			if visited[i] == 0:
				d1[parent].append(i)
				stack.append((i,temp))

	return d1
for i in range(n-1):

	a,b = map(int,input().split())
	d[a].append(b)
	d[b].append(a)
parent = 0

visited = [0 for i in range(n+1)]


x = 1

state = list(map(int,input().split()))
goal = list(map(int,input().split()))
c = None
x = 1
parent = 0
d1 = DFS(d,x,parent,visited)

visited = [0 for i in range(n+1)]
l = []
ans = DFSop(d1,x,goal,state,c,visited,l)

for i in d[1]:
	d1 = DFS(d,x,parent,visited)
	c = None

	ans = ans + DFSop(d1,i,goal,state,c,visited,l)
ans = set(ans)
print(len(ans))
for i in ans:
	print(i)
