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).
# /usr/bin/env python3.8
# -*- coding:utf-8 -*-
# author:guguanghui
n = int(input())
adj = list([] for i in range(n+1))
for _ in range(n - 1):
    u, v = map(int, input().split())
    adj[u].append(v)
    adj[v].append(u)
a = [0] + list(map(int, input().split()))
b = [0] + list(map(int, input().split()))
def dfs(u, p, c_lvl, p_lvl, d):
    stk = [(u, p, c_lvl, p_lvl)]
    while stk:
        u, p, c_lvl, p_lvl = stk.pop()
        if c_lvl != d[u]:
            c_lvl = 1 - c_lvl
            res.append(str(u))
        for v in adj[u]:
            if v != p:
                stk += [(v, u, p_lvl, c_lvl)]

d = [i ^ j for (i, j) in zip(a, b)]
res = []
dfs(1, 0, 0, 0, d)
print(len(res))
print('\n'.join(res))