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*<sub class="lower-index">*i*</sub>, 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*<sub class="lower-index">*i*</sub>, which can also be only 0 or 1. You need to reach the goal of the game by using minimum number of operations.
The first line contains an integer *n* (1 ≤ *n* ≤ 10<sup class="upper-index">5</sup>). Each of the next *n* - 1 lines contains two integers *u*<sub class="lower-index">*i*</sub> and *v*<sub class="lower-index">*i*</sub> (1 ≤ *u*<sub class="lower-index">*i*</sub>, *v*<sub class="lower-index">*i*</sub> ≤ *n*; *u*<sub class="lower-index">*i*</sub> ≠ *v*<sub class="lower-index">*i*</sub>) meaning there is an edge between nodes *u*<sub class="lower-index">*i*</sub> and *v*<sub class="lower-index">*i*</sub>. 

The next line contains *n* integer numbers, the *i*-th of them corresponds to *init*<sub class="lower-index">*i*</sub> (*init*<sub class="lower-index">*i*</sub> is either 0 or 1). The following line also contains *n* integer numbers, the *i*-th number corresponds to *goal*<sub class="lower-index">*i*</sub> (*goal*<sub class="lower-index">*i*</sub> is either 0 or 1).
#include <iostream>
#include <vector>
#include <unordered_map>
using namespace std;

unordered_map<int, vector<int>> graph;
vector<int> clr;
vector<int> wnt_clr;
int mn_op = 0;
vector<int> ans;

void dfs(int u, int parent, int level, int even_level, int odd_level) {
    if (level % 2 == 0) {
        if (even_level % 2) {
            if (clr[u - 1] == wnt_clr[u - 1]) {
                mn_op++;
                even_level++;
                ans.push_back(u);
            }
        } else {
            if (clr[u - 1] != wnt_clr[u - 1]) {
                mn_op++;
                even_level++;
                ans.push_back(u);
            }
        }
    } else {
        if (odd_level % 2) {
            if (clr[u - 1] == wnt_clr[u - 1]) {
                mn_op++;
                odd_level++;
                ans.push_back(u);
            }
        } else {
            if (clr[u - 1] != wnt_clr[u - 1]) {
                mn_op++;
                odd_level++;
                ans.push_back(u);
            }
        }
    }

    for (int v : graph[u]) {
        if (v != parent) {
            dfs(v, u, level + 1, even_level, odd_level);
        }
    }
}

int main() {
    int n;
    cin >> n;

    for (int i = 0; i < n - 1; i++) {
        int u, v;
        cin >> u >> v;
        graph[u].push_back(v);
        graph[v].push_back(u);
    }

    clr.resize(n);
    wnt_clr.resize(n);
    for (int i = 0; i < n; i++) {
        cin >> clr[i];
    }
    for (int i = 0; i < n; i++) {
        cin >> wnt_clr[i];
    }

    dfs(1, 0, 0, 0, 0);

    cout << mn_op << endl;
    for (int val : ans) {
        cout << val << endl;
    }

    return 0;
}