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 sys
input=sys.stdin.readline 
n=int(input())
g=[[] for i in range(n+1)]
vis=[0 for i in range(n+1)]
par=[0 for i in range(n+1)]
inv=[0 for i in range(n+1)]
ans=[]

def dfs(u):
    vis[u]+=1
    for v in g[u]:
        if vis[v]==0:
            par[v]=u
            dfs(v)

def dfs2(u):
    vis[u]+=1
    if u==1 or par[u]==1:
        if cur[u-1]!=goal[u-1]:
            inv[u]+=1
            ans.append(u)
    else :
        inv[u]=inv[par[par[u]]]
        if inv[par[par[u]]]%2:
            cur[u-1]^=1
        if cur[u-1]!=goal[u-1]:
            inv[u]+=1
            ans.append(u)
    for v in g[u]:
        if vis[v]==0:
            dfs2(v)


for i in range(n-1):
    u,v=map(int,input().split())
    g[u].append(v)
    g[v].append(u)

dfs(1)

cur=list(map(int,input().split()))
goal=list(map(int,input().split()))

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

dfs2(1)

print(len(ans))
for j in ans:
    print(j)
