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).
import itertools as it
import collections as cc
import heapq as hp
import sys
I=lambda : list(map(int,input().split()))
import operator as op
from functools import reduce
n,=I()
g=[[] for i in range(n)]
for i in range(n-1):
	x,y=I()
	x-=1;y-=1
	g[x].append(y)
	g[y].append(x)
a=I()
b=I()
p=[-1]*n 
visi=[0]*n
te=[]
def dfs(u,pa,l,od,ev):
	if l%2:
		if a[u]^od!=b[u]:
			od^=1
			te.append(u+1)
	else:
		if a[u]^ev!=b[u]:
			ev^=1
			te.append(u+1)
	for v in g[u]:
		if v==pa:
			continue
		dfs(v,u,l+1,od,ev)
dfs(0,0,1,0,0)
ans=0
print(len(te))
print(*te)