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 *
setrecursionlimit(10 ** 6)
n = int(input())
graph = [[] for i in range(n + 1)]
for i in range(n-1):
	u , v = list(map(int,input().split()))
	graph[u].append(v)
	graph[v].append(u)
a = [-1] + list(map(int,input().split()))
b = [-1] + list(map(int,input().split()))
dp = [0 , 1]
vis = [0 for i in range(n+1)]
count = 0
out = []
def dfs(u):
	global count
	vis[u] = 1
	if dp[a[u]] != dp[b[u]]:
		count += 1
		dp[0] = (dp[0] + 1) % 2
		dp[1] = (dp[1] + 1) % 2
		out.append(u)
	for i in graph[u]:
		if vis[i] == 0:
			dfs(i)
dfs(1)
print(count)
for i in out:
	print(i)