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).
n=int(input())
from collections import defaultdict
graph=defaultdict(list)
for i in range(n-1):
    u,v=[int(i) for i in input().split()]
    graph[u].append(v)
    graph[v].append(u)

curr=[int(i) for i in input().split()]
goal=[int(i) for i in input().split()]

ops=[]
def dfs(parent,ver,odd,even,lev):
    par=lev%2
    temp=curr[ver-1]
    changed=0
    if par and odd%2:
        temp=curr[ver-1]^1
    if par==0 and even%2:
        temp=curr[ver-1]^1
    #print(curr[ver-1],goal[ver-1])
    if temp!=goal[ver-1]:
        ops.append(ver)
        changed=1
    

    for ne in graph[ver]:
        if ne!=parent:
            if changed:
                if lev%2:
                    dfs(ver,ne,odd+1,even,lev+1)
                else:
                    dfs(ver,ne,odd,even+1,lev+1)
            else:
                dfs(ver,ne,odd,even,lev+1)

dfs(-1,1,0,0,0)
print(len(ops))
for i in ops:
    print(i)