CodeForces - 429A Xor-tree

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_i, 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_i, 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 u_i and v_i (1 ≤ u_i, v_i ≤ n; u_i ≠ v_i) meaning there is an edge between nodes u_i and v_i.

The next line contains n integer numbers, the i-th of them corresponds to init_i(init_i is either 0 or 1). The following line also contains n integer numbers, the i-th number corresponds to goal_i (goal_i 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 x_i, representing that you pick a node x_i.

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]);
}