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 sys import stdin,stdout
n=int(input())
a=[[] for i in range(n)]
for i in range(n-1):
    c,d=(int(o) for o in  stdin.readline().split())
    a[c-1].append(d-1);a[d-1].append(c-1)
b=list(int(o) for o in  stdin.readline().split())
c=list(int(o) for o in  stdin.readline().split())
lv=[0]*n;lv[0]=1
q=[0]
while len(q)!=0:
    r=q.pop()
    for i in a[r]:
        if lv[i]==0:
            lv[i]=lv[r]+1
            q.append(i)
ans='';vis=[True]*n
vis[0]=False;k=0
if b[0]==c[0]:
    q=[(0,0,0)]
else:
    q=[(0,1,0)];ans+=('1'+'\n');k+=1
while len(q)!=0:
    no,o,e=q.pop()
    for i in a[no]:
        if vis[i]:
            l=e;m=o
            vis[i]=False
            if lv[i]%2==0:
                if b[i]==c[i] and l%2!=0:
                    ans+=(str(i+1)+'\n')
                    l+=1;k+=1
                elif b[i]!=c[i] and l%2==0:
                    ans+=(str(i+1)+'\n')
                    l+=1;k+=1
            else:
                if b[i]==c[i] and m%2!=0:
                    ans+=(str(i+1)+'\n')
                    m+=1;k+=1
                elif b[i]!=c[i] and m%2==0:
                    ans+=(str(i+1)+'\n')
                    m+=1;k+=1
            q.append((i,m,l))
print(k)
stdout.write(ans)