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
sys.setrecursionlimit(int(1e4))

def solve(cur, ooe, lvl, flipodd, flipeven, parent):
    if init[cur] != goal[cur]:
        if ooe == 'odd':
            if flipodd is True:
                for child in tree[cur]:
                    if child != parent:
                        solve(child, 'even', lvl + 1, True, flipeven, cur)
            else:
                new.append(cur)
                for child in tree[cur]:
                    if child != parent:
                        solve(child, 'even', lvl + 1, True, flipeven, cur)
            
        elif ooe == 'even':
            if flipeven is True:
                for child in tree[cur]:
                    if child != parent:
                        solve(child, 'odd', lvl + 1, flipodd, True, cur)
            else:
                new.append(cur)
                for child in tree[cur]:
                    if child != parent:
                        solve(child, 'odd', lvl + 1, flipodd, True, cur)
                        
    elif init[cur] == goal[cur]:
        if ooe == 'odd':
            if flipodd is True:
                new.append(cur)
                for child in tree[cur]:
                    if child != parent:
                        solve(child, 'even', lvl + 1, False, flipeven, cur)
            else:
                for child in tree[cur]:
                    if child != parent:
                        solve(child, 'even', lvl + 1, False, flipeven, cur)
        elif ooe == 'even':
            if flipeven is True:
                new.append(cur)
                for child in tree[cur]:
                    if child != parent:
                        solve(child, 'odd', lvl + 1, flipodd, False, cur)
            else:
                for child in tree[cur]:
                    if child != parent:
                        solve(child, 'odd', lvl + 1, flipodd, False, cur)

n = int(input())
from collections import defaultdict
tree = defaultdict(list)
for j in range(n - 1):
    u, v = map(int, input(). split())
    tree[u].append(v) ; tree[v].append(u)
init = [0] + list(map(int, input(). split()))
goal = [0] + list(map(int, input(). split()))
new = []
solve(1, 'odd', 1, False, False, None)
print(len(new))
for i in new:
    print(i)