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 os
import sys

sys.setrecursionlimit(10 ** 5)
from io import BytesIO, IOBase
from collections import Counter, defaultdict
import math
import heapq
import bisect
import collections


def ceil(a, b):
    return (a + b - 1) // b


BUFSIZE = 8192
inf = float('inf')


def get():
    return sys.stdin.readline()


mod = 10 ** 5 + 7
import collections,sys,threading
sys.setrecursionlimit(10**9)
threading.stack_size(10**8)

# for _ in range(int(get())):
# n=int(get())
# l=list(map(int,get().split()))
# = map(int,get().split())
##################################################################################
def solve():
    def dfs(nums, prev, odd_time, even_time, level):
        if level % 2 == 1:
            cur = (a[nums] + odd_time) % 2
        else:
            cur = (a[nums] + even_time) % 2
        if cur != b[nums]:
            if level % 2 == 1:
                odd_time += 1
            else:
                even_time += 1
            result.append(nums)
        for i in d[nums]:
            if i == prev:
                continue
            dfs(i, nums, odd_time, even_time, level + 1)

    ##################################################################################
    n = int(get())
    result = []
    d = [[] for i in range(n + 1)]
    for i in range(n - 1):
        u, v = map(int, get().split())
        d[u].append(v)
        d[v].append(u)
    a = [0] + list(map(int, get().split()))
    b = [0] + list(map(int, get().split()))
    dfs(1, -1, 0, 0, 0)
    print(len(result))
    for i in result:
        print(i)

threading.Thread(target=solve).start()