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
from functools import lru_cache, cmp_to_key
from heapq import merge, heapify, heappop, heappush
from math import *
from collections import defaultdict as dd, deque, Counter as C
from itertools import combinations as comb, permutations as perm
from bisect import bisect_left as bl, bisect_right as br, bisect, insort
from time import perf_counter
from fractions import Fraction
import copy
from copy import deepcopy
import time
starttime = time.time()
mod = int(pow(10, 9) + 7)
mod2 = 998244353

def data(): return sys.stdin.readline().strip()
def out(*var, end="\n"): sys.stdout.write(' '.join(map(str, var))+end)
def L(): return list(sp())
def sl(): return list(ssp())
def sp(): return map(int, data().split())
def ssp(): return map(str, data().split())
def l1d(n, val=0): return [val for i in range(n)]
def l2d(n, m, val=0): return [l1d(n, val) for j in range(m)]
try:
    # sys.setrecursionlimit(int(pow(10,6)))
    sys.stdin = open("input.txt", "r")
    # sys.stdout = open("output.txt", "w")
except:
    pass
def pmat(A):
    for ele in A: print(*ele,end="\n")

# from sys import stdin
# input = stdin.buffer.readline
# I = lambda : list(map(int,input().split()))

n=L()[0]
g=[[] for i in range(n+1)]
for _ in range(n-1):
    x,y=L()
    g[x].append(y)
    g[y].append(x)

global ans 
ans=[]
A=[0]+L()
B=[0]+L()
vis=[0 for i in  range(n+1)]
def dfs(i,para,parb):
    vis[i]=1
    if (A[i]+parb)%2==B[i]:
        for ele in g[i]:
            if vis[ele]==0:
                dfs(ele,parb,para)
    else:
        ans.append(i)
        for ele in g[i]:
            if vis[ele]==0:
                dfs(ele,(parb+1)%2,para)


dfs(1,0,0)
print(len(ans))
for ele in ans:
    print(ele)