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
sys.setrecursionlimit(10**6);

# sys.stdin = open('input.txt', 'r')
# sys.stdout = open('output.txt', 'w')




n=int(input());
a=[[] for i in range(n+1)];vis=[0]*(n+1);
for i in range(n-1):
    u,v=map(int,input().split());
    a[u].append(v);a[v].append(u);
curr=[-1]+list(map(int,input().split()));
goal=[-1]+list(map(int,input().split()));
ans=[];
def dfs(i,lvl,o,e):
    c=0;
    if(lvl%2==0):c=(curr[i]+e)%2;
    else:c=(curr[i]+o)%2;
    if(c!=goal[i]):
        if(lvl%2==0):e+=1;
        else:o+=1;
        ans.append(i);
    for j in a[i]:
        if(not vis[j]):
            vis[j]=1;
            dfs(j,lvl+1,o,e);
dfs(1,0,0,0);
print(len(ans));
for i in ans:
    print(i);
