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).
n = int(input())
tree = {}
for i in range(n-1):
    a,b = [int(x) for x in input().split()]
    if a not in tree: tree[a] = []
    if b not in tree: tree[b] = []
    tree[a].append(b)
    tree[b].append(a)
numb = [int(x) for x in input().split()]
mubiao = [int(x) for x in input().split()]
flipp = [0]*n
shangyige = [0]*n

queue = [1]
visited = set()
count = 0
printt = []

def flip(x):
    shuchu = 0
    t = 0
    i = shangyige[x-1]
    while 1:
        if i == 0: break
        else:
            shuchu += t*flipp[i-1]
            t = 1-t
            i = shangyige[i-1]
    return shuchu

while len(queue)>0:
    x = queue.pop(0)
    visited.add(x)
    numb[x-1] = (numb[x-1] + flip(x)) % 2
    if numb[x-1] != mubiao[x-1]:
        flipp[x-1] = 1
        numb[x-1] = 1 - numb[x-1]
        count += 1
        printt.append(x)
    for k in tree[x]:
        if k not in visited:
            queue.append(k)
            flag = True
        else: shangyige[k-1] = x
print(count)
for x in printt: print(x)