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.io.OutputStream;
import java.io.IOException;
import java.io.InputStream;
import java.io.PrintWriter;
import java.util.StringTokenizer;
import java.io.IOException;
import java.io.BufferedReader;
import java.util.LinkedList;
import java.io.InputStreamReader;
import java.io.InputStream;

/**
 * Built using CHelper plug-in
 * Actual solution is at the top
 */
public class Main {
	public static void main (String[] args) {
		InputStream inputStream = System.in;
		OutputStream outputStream = System.out;
		InputReader in = new InputReader (inputStream);
		PrintWriter out = new PrintWriter (outputStream);
		TaskA solver = new TaskA ();
		solver.solve (1, in, out);
		out.close ();
	}
	
	static class TaskA {
		private static LinkedList<Integer>[] adlst;
		private static LinkedList<Integer> ans;
		private static byte[] vstd;
		private static int cnt;
		private static byte flip;
		private static byte flg;
		private static int[] init;
		private static int[] goal;
		
		private void dfsrcrsv (int op) {
			vstd[op] = 1;
			byte tflip = flip, tflg = flg;
			if (flip != 0) {
				init[op] ^= 1;
			}
//		System.out.println (op+" "+flip+flg+" "+init[op]);
			if (init[op] != goal[op]) {
				cnt++;
				ans.add (op);
				init[op] ^= 1;
/*now if I see that flip is one then I will have to make flip 0 but if flip is 0 but flag is one then flip is two*/
				if (flip == 1) {
					flg = 0;
					flip = 0;
				}
				else if (flip == 2) {
					flip = 0;//ekhon theke eita ar flip kora zabe na
				}
				else if (flg == 1) {
					flip = 2;
				}
				else if (flg == 0) {
					flg = 1;
					flip = 1;
				}
				
			}
			if (flg == 1 && flip != 2) {
				flip ^= 1;
			}
/*flip er man maybe more than two hoite pare, as zodi porpor duita node flip kore then everyone is going to flip*/
			for (int os : adlst[op]) {
				if (vstd[os] == 0) {
					dfsrcrsv (os);
				}
			}
			flip = tflip;
			flg = tflg;
		}
		
		public void solve (int testNumber, InputReader in, PrintWriter out) {
/*Tree is a connected component so no need to worry about it, now the worry is whether to use dfs or bfs.
As it is a tree so I think we should use dfs, now have to think whether tree is directed or not, I think it's directed
but shit man, how to implement the fact that when root is flipped all the childs are flipped or not flipped.
I think I should read the article about lazy propagation in shafaet blog as I have forgotten the trick
Well, I can do one thing that is to mark whether any of the ancestor of the node has been flipped or not, if yes then
use variable flip to determine whether to flip it or not, by the even flag can be changed.*/
			int n = in.nextInt (), u, v;
			cnt = flip = flg = 0;
			adlst = new LinkedList[n+5];
			vstd = new byte[n+5];
			ans = new LinkedList<> ();
			boolean[] taken = new boolean[n+5];
			taken[1] = true;
			for (int i = 0; i<n+5; i++) {
				adlst[i] = new LinkedList<> ();
			}
			for (int i = 1; i<n; i++) {
				u = in.nextInt ();
				v = in.nextInt ();
				if (taken[u]) {
					adlst[u].add (v);
					taken[v] = true;
				}
				else {
					adlst[v].add (u);
					taken[u] = true;
				}
			}
			init = new int[n+5];
			goal = new int[n+5];
			for (int i = 1; i<=n; i++) {
				init[i] = in.nextInt ();
			}
			for (int i = 1; i<=n; i++) {
				goal[i] = in.nextInt ();
			}
//it is clearly told that the root of the tree is node 1, so obviously dfs is going to start from root
			dfsrcrsv (1);
			out.println (cnt);
			for (int i : ans) {
				out.println (i);
			}
		/*for (int i : init) {
			out.print (i+" ");
		}out.println();*/
		}
		
	}
	
	static class InputReader {
		private BufferedReader reader;
		private StringTokenizer tokenizer;
		
		public InputReader (InputStream stream) {
			reader = new BufferedReader (new InputStreamReader (stream), 32768);
			tokenizer = null;
		}
		
		public String next () {
			while (tokenizer == null || !tokenizer.hasMoreTokens ()) {
				try {
					String str;
					if ((str = reader.readLine ()) != null) tokenizer = new StringTokenizer (str);
					else return null;//to detect eof
				} catch (IOException e) {
					throw new RuntimeException (e);
				}
			}
			return tokenizer.nextToken ();
		}
		
		public int nextInt () {
			return Integer.parseInt (next ());
		}
		
	}
}