/*

考虑直接使用暴力来算的话   SG[i]表示以i为根的子树的SG值， 然后考虑枚举删除那个子树节点， 然后求拆成的树的sg异或值， 求mex即可 复杂度三次方

然后考虑尝试 整体来做 发现对于每次子树的合并， 每棵子树种的sg值相当是异或了其他的所有子树

而这个东西显然是可以用 线段树合并来维护的

最后找出所有sg不为0的点即可

*/

#include<cstdio>

#include<algorithm>

#include<cstring>

#include<iostream>

#include<queue>

#include<cmath>

#define ll long long

#define M 100010

#define mmp make_pair

using namespace std;

int read() {

	int nm = 0, f = 1;

	char c = getchar();

	for(; !isdigit(c); c = getchar()) if(c == '-') f = -1;

	for(; isdigit(c); c = getchar()) nm = nm * 10 + c - '0';

	return nm * f;

}

int cor[M], cov[M * 60], laz[M * 60], ls[M * 60], rs[M * 60], rt[M], n, sg[M], cnt, ans[M], tot;

vector<int> to[M];

void add(int &now, int v, int deep) {

	if(deep <= -1) return;

	if(v & (1 << deep)) swap(ls[now], rs[now]);

	laz[now] ^= v;

}

void pushdown(int now, int deep) {

	if(!laz[now] || !now) return;

	add(ls[now], laz[now], deep - 1);

	add(rs[now], laz[now], deep - 1);

	laz[now] = 0;

}

int merge(int now, int last, int deep) {

	if(!now || !last) return now | last;

	if(deep == -1) {

		cov[now] |= cov[last];

		return now;

	}

	pushdown(now, deep), pushdown(last, deep);

	ls[now] = merge(ls[now], ls[last], deep - 1);

	rs[now] = merge(rs[now], rs[last], deep - 1);

	cov[now] = cov[ls[now]] && cov[rs[now]];

	return now;

}

void insert(int &now, int v, int deep) {

	if(!now) now = ++cnt;

	if(deep == -1) {

		cov[now] = 1;

		return;

	}

	if(v & (1 << deep)) insert(rs[now], v, deep - 1);

	else insert(ls[now], v, deep - 1);

}

int mex(int now, int deep) {

	if(!now || deep == -1) return 0;

	pushdown(now, deep);

	if(!cov[ls[now]]) return mex(ls[now], deep - 1);

	else return (1 << deep) + mex(rs[now], deep - 1);

}

void dfs(int now, int fa) {

	int tmp = 0;

	for(int i = 0; i < to[now].size(); i++) {

		int vj = to[now][i];

		if(vj == fa) continue;

		dfs(vj, now);

		tmp ^= sg[vj];

	}

	if(!cor[now]) insert(rt[now], tmp, 17);

	for(int i = 0; i < to[now].size(); i++) {

		int vj = to[now][i];

		if(vj == fa) continue;

		add(rt[vj], tmp ^ sg[vj], 17);

		rt[now] = merge(rt[now], rt[vj], 17);

	}

	sg[now] = mex(rt[now], 17);

}

void get(int now, int fa, int anss) {

	for(int i = 0; i < to[now].size(); i++) {

		int vj = to[now][i];

		if(vj == fa) continue;

		anss ^= sg[vj];

	}

	if(anss == 0 && !cor[now]) ans[++tot] = now;

	for(int i = 0; i < to[now].size(); i++) {

		int vj = to[now][i];

		if(vj == fa) continue;

		get(vj, now, anss ^ sg[vj]);

	}

}

int main() {

	n = read();

	for(int i = 1; i <= n; i++) cor[i] = read();

	for(int i = 1; i < n; i++) {

		int vi = read(), vj = read();

		to[vi].push_back(vj);

		to[vj].push_back(vi);

	}

	dfs(1, 0);

	get(1, 0, 0);

	if(tot) {

		sort(ans + 1, ans + tot + 1);

		for(int i = 1; i <= tot; i++) cout << ans[i] << '\n';

	} else puts("-1");

	return 0;

}