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 initi, 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 goali, which can also be only 0 or 1. You need to reach the goal of the game by using minimum number of operations.

Input

The first line contains an integer n (1 ≤ n ≤ 105). Each of the next n - 1 lines contains two integers ui and vi (1 ≤ ui, vi ≤ n; ui ≠ vi) meaning there is an edge between nodes ui and vi.

The next line contains n integer numbers, the i-th of them corresponds to initi(initi is either 0 or 1). The following line also contains n integer numbers, the i-th number corresponds to goali (goali is either 0 or 1).

Output

In the first line output an integer number cnt, representing the minimal number of operations you perform. Each of the next cnt lines should contain an integer xi, representing that you pick a node xi.

Examples

Input
10
2 1
3 1
4 2
5 1
6 2
7 5
8 6
9 8
10 5
1 0 1 1 0 1 0 1 0 1
1 0 1 0 0 1 1 1 0 1
Output
2
4
7

题意：给出一棵树，然后这个树，然后每个节点有个值，然后再给一个目标树，要求把树翻转成目标树，翻转规则是，翻转当前节点，他的儿子不变
，他的儿子的儿子要翻转，求翻转次数

思路：因为他每翻一个下面的节点就要发生变化，翻多了节点后就不好判断当前要不要翻转，我就分别记录，奇数层的翻转次数，
和偶数层的翻转次数还有就是判断当前层的奇偶，然后进行dfs
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
vector<int> mp[100001];
int c1[100001],c2[100001];
int n,x,y;
int num[100001];
int vis[100001];
int cnt;
void dfs(int x,int j,int o,int z,int d)
{
    if(z==0)//第一个不同的点
    {
        if(c1[x]!=c2[x])
        {
            num[cnt++]=x;
            d=1;
            z=1;
            j=1; 
        }
    }
    else{//判断奇偶与层数
        d++;
        if(c1[x]==c2[x]&&d%2==1&&j%2==1) 
        {
            num[cnt++]=x;
            j++;
        }
        else if(c1[x]!=c2[x]&&d%2==1&&j%2==0) 
        {
            num[cnt++]=x;
            j++;
        }
        else if(c1[x]==c2[x]&&d%2==0&&o%2==1) 
        {
            num[cnt++]=x;
            o++;
        }
        else if(c1[x]!=c2[x]&&d%2==0&&o%2==0) 
        {
            num[cnt++]=x;
            o++;
        }
        
    }
    for(int i=0;i<mp[x].size();i++)//往下dfs
    {
        if(vis[mp[x][i]]==0)
        {
            vis[mp[x][i]]=1;
            dfs(mp[x][i],j,o,z,d);
        }
    }
} 
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n-1;i++)//领接表建图
    {
        scanf("%d%d",&x,&y);
        mp[x].push_back(y);
        mp[y].push_back(x);
    }
    for(int i=1;i<=n;i++)
        scanf("%d",&c1[i]);
    for(int i=1;i<=n;i++)
        scanf("%d",&c2[i]);
    vis[1]=1;
    dfs(1,0,0,0,0);
    printf("%d\n",cnt);
    for(int i=0;i<cnt;i++)
        printf("%d\n",num[i]);
}