Problem: 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

Solution (in JAVA):

Solution:
import java.util.*;
/** 
 * Johan Sebastián Gómez Luna - 02286062
 * Data Structures - UN
 **/

public class XorTree {
    
    private static int size;
    private static ArrayList[] key;
    private static int start[] ;
    private static int finish[] ;
    private static int klm;
    private static ArrayList <Integer> moves;
    
    public static void bfs(int tony, int howard, int mark1, int mark2){
       
       int stark = howard;
       int c1 = mark1;
       int c2 = mark2;
       //Se aplican los "flip", o cambios asociados a cada uno de los nodos para dar con la solucion final.
       //Se tiene en cuanta la repercicion del "flip" en un padre, sobre sus nietos.
       //Si el muero asociado es 1, el nodo no es modificado, pero si por el contrario es 0, se realiza el "flip".
       if(c2==1){
          start[tony]=flip(start[tony]); 
       }
       if(c2==0){
           
           if(start[tony]==finish[tony]){
               c2=c1;
               c1=0;
           }
           else
           {
               c2=c1;
               c1=1;
               start[tony]=flip(start[tony]);
               klm++;
               moves.add(tony);
           }
       }
       else
       {
           if(start[tony]==finish[tony]){
               c2=c1;
               c1=1;
           }
           else{
               c2=c1;
               c1=0;
               start[tony]=flip(start[tony]);
               klm++;
               moves.add(tony);
           }
       }
       //Se asegura de que exista al menos un hijo que visitar
       if(key[tony]!= null && key[tony].isEmpty()!= true){
           while(key[tony].isEmpty()!=true){
                int ironMan = (int)key[tony].remove(0);
                    if(ironMan==stark){
                    }
                    else{
                        bfs(ironMan,tony,c1,c2);
                    }
           }
       } 
    }
    
    public static int flip (int i){
        if (i==0)
            return 1;
        else
            return 0;
    }
    
    public static void main(String[] args) {
        //Input. Se realiza la recepcion de datos para cada caso.
        Scanner sc = new Scanner( System.in );
		size = sc.nextInt();
                int u,v;
                
                //Alamacena las posiciones de cada hijo, haciendo referencia al padre.
                key = new ArrayList[size+1];
                
                start = new int [size+1];
                finish = new int [size+1];
                
		for(int i=1 ; i < size ; i++){
                       //Se asignan los valosres de "u" y "v"
                       u = sc.nextInt();
                       v = sc.nextInt();
                       
                       //Se agregan los hijos al arreglo.
                            if(key[u]==null){  //Si no existe la lista en dicha posicion, se crea una.
                                 key[u]= new ArrayList <Integer>();
                            }
                            if(key[v]==null){
                                 key[v]= new ArrayList <Integer>();
                            }
                            key[u].add(v);
                            key[v].add(u);
            }
                
            for(int i=1; i <= size; i++){
                start[i]= sc.nextInt();
            } 
            for(int i=1; i <= size; i++){
                finish[i]= sc.nextInt();
            } 
            //Se crea una lista con los nodos que han sido modificados, para luego determinar la solucion.
            moves = new ArrayList <Integer>(); 
            
        //Se inicializa la BFS. Como no tiene padre, el segundo campo es 0, y como no tiene ni padre ni abuelo,
        //no se pueden registrar los cambios, asi que su valor tambien sera 0.
        bfs(1,0,0,0);
        //Impirme el numero de nodos modificados para llegar a la solucion.
        System.out.println(klm);
        for(int i=0; i < moves.size(); i++){
            System.out.println(moves.get(i));
        }
    }
}