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
d = defaultdict(list)
for i in range(n-1):
    a,b = [int(i) for i in input().split()]
    d[a].append(b)
    d[b].append(a)
a  =[int(i) for i in input().split()]
b = [int(i) for i in input().split()]
visited = [False for i in range(n+1)]
def dfs(v1,level,even,odd):
    count=0
    stack=[]
    res=[]
    stack.append((v1,level,even,odd))
    while(stack):
        v1,level,even,odd =stack.pop()
        visited[v1]=True
        if level&1==0:
            if even&1!=0:
                if a[v1-1]!=b[v1-1]:
                    a[v1-1]^1
            if a[v1-1]!=b[v1-1]:
                a[v1-1]^1
                count+=1
                res.append(v1)
                even+=1
        else:
            if odd&1!=0:
                if a[v1-1]!=b[v1-1]:
                    a[v1-1]^1
            if a[v1-1]!=b[v1-1]:
                a[v1-1]^1
                res.append(v1)
                count+=1
                odd+=1
        for i in d[v1]:
            if visited[i]==False:
                stack.append((i,level+1,even,odd))
    return count,res
c,res=dfs(1,0,0,0)
print(c)
for i in res[::-1]:
    print(i)
        