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).
def dfs(n,d,a,b):
    q = [1]
    visited = [1] 
    d1,d2 = {} ,{}
    
    for i in range(1,n+1):
        d1[i] = 0
        d2[i] = 0 
    
    n = 1
    i = 0 
    count = 0 
    lst = [] 
    while i<n:
        m = q[i]
        if d1[m]%2==0:
            if a[m]==b[m]:
                count += 0 
            else:
                count+=1
                d1[m]+=1 
                lst.append(m)
        else:
            if a[m]!=b[m]:
                count += 0 
            else:
                count+=1
                d1[m]+=1 
                lst.append(m)
        
        for j in d[m]:
            if j not in visited:
                visited.append(j)
                
                d2[j] = d1[m] 
                
                q.append(j)
                n+=1
        i+= 1
    return (count,lst)


n = int(input())
d = {} 
for i in range(1,n+1):
    d[i] = [] 
for _ in range(n-1):
    x,y = map(int,input().split())
    d[x].append(y)
    d[y].append(x)
    
a = [0]+list(map(int,input().split()))
b = [0]+ list(map(int,input().split()))

count, lst = dfs(n,d,a,b)
print(count)
for i in lst:
    print(i)
    
    