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:

/**
 *
 * @author Alejandro
 */


import java.util.ArrayList;
import java.util.List;
import java.util.Scanner;
import java.util.Stack;



public class Taller3c {
    
    // Se crea un arreglo de arreglos.
     static List<Integer>[] parents; // arreglo que contiene la relacion entre nodos, ya sea de padre o hijos
     static List<Integer> changes; // Lista donde se va almacenar los valores que hay que flipear
     static int[] init, goal; // arreglos donde se almacenan los valores inicial y final
     static int size;        //number of nodes

       static class Node {
       // el nodo contiene 4 estados, el primero es el nodo especifico, , el segundo es el padre del nodo, y los dos ultimos
       // son banderas que permiten saber si los ancestros de dos en dos ya sean pares o impares han sido cambiados
        public int node, parent, imp, par;
        
        // Constructor
        public Node(int node, int parent, int imp, int par) {
            this.node = node;
            this.parent = parent;
            this.imp = imp;
            this.par = par;
        }
    }
    

    public static void main(String[] args) {

        //Escanea la entrada e inicializa las matrices de entrada y salida
        Scanner sc = new Scanner(System.in);
        size = sc.nextInt();
        parents = new List[size + 1];
        init = new int[size + 1];
        goal = new int[size + 1];
        
               for (int i = 0; i < parents.length; i++) {
            parents[i] = new ArrayList<>(); // crea un arreglo de arreglos
        }
     
        changes = new ArrayList<>();
      //Agrega al arreglo todos los nodos relacionados, padres hijos o hijos padres
        for (int i = 1; i <= size - 1; i++) {
            int pi = sc.nextInt();
            int ip = sc.nextInt();

            parents[pi].add(ip); 
            parents[ip].add(pi);
        }
      
        for (int i = 1; i <= size; i++) {
            init[i] = sc.nextInt(); //recibe el arreglo entrada
        }
        // LLena el arreglo de salida
        for (int i = 1; i <= size; i++) {
            goal[i] = sc.nextInt();
        }       
        
        //  Funciona como una dfs crea una pila
        Stack<Node> stack = new Stack<>();
        // Primer estado
        Node c= new Node(1, -1, 0, 0);

        stack.push(c);
        while (!stack.isEmpty()) {
            c= stack.pop();
           
            if ((init[c.node] != goal[c.node] && c.imp == 0)
                    || (init[c.node] == goal[c.node] && c.imp == 1)) {
                //  Si el nodo es diferene al de la meta hace un flip
                c.imp ^= 1;
               
                changes.add(c.node);
            }
      
            for (int i = 0; i < (int) parents[c.node].size(); i++) {
              
                if (c.parent == parents[c.node].get(i)) {
                    continue;
                }
           
                stack.push(new Node(parents[c.node].get(i), c.node, c.par, c.imp));
            }
        }
        
        // Imprime los nodos que hay que flipear
        int s = changes.size();
        System.out.println(s);
        int i=0;
        for(int ii: changes)
          {
            System.out.println(changes.get(i++));
        }
    }
}