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
input=sys.stdin.readline
from collections import deque
n=int(input())
a=[[] for i in range(n+1)]
oc=0
ec=0
for i in range(n-1):
    x,y=map(int,input().split())
    a[x].append(y)
    a[y].append(x)
p=[int(x) for x in input().split()]    
q=[int(x) for x in input().split()]   
r=[0]
for i in range(n):
    if p[i]==q[i]:
        r.append(0)
    else:
        r.append(1)
vis=[0 for i in range(n+1)]
c=0
l=[0 for i in range(n+1)]
oc=[0 for i in range(n+1)]
ec=[0 for i in range(n+1)]
#print(r)
def bfs(c,oc,ec):
    
    q=deque()
    q.append(1)
    vis[1]=1
    if r[i]:
        ec[i]+=1
        c+=1
    while len(q)>0:
        z=q.popleft()
        
        for j in a[z]:
            if vis[j]==0:
                #print(oc,ec,l[j],l[z],j,z)
                l[j]=l[z]^1
                if l[j]:
                    if oc[z]%2!=r[j]&1:
                        oc[j]+=1
                        c+=1
                else:
                    if ec[z]&1!=r[j]&1:
                        ec[j]+=1
                        c+=1
                vis[j]=1
                q.append(j)
                #print(oc,ec,l[j],l[z],j,z,c)
    return c        
print(bfs(c,oc,ec))