#include <bits/stdc++.h>
#include "atcoder/all"
#define debug cout << "OK" << endl;
template<typename T>
inline bool chmax(T& a, const T b) { if (a < b) { a = b; return true; } return false; }
template<typename T>
inline bool chmin(T& a, const T b) { if (a > b) { a = b; return true; } return false; }
inline int Code(char c) {
	if ('A' <= c and c <= 'Z')return (int)(c - 'A');
	if ('a' <= c and c <= 'z')return (int)(c - 'a');
	if ('0' <= c and c <= '9')return (int)(c - '0');
	assert(false);
}
using namespace std;
using namespace atcoder;
#define int long long
constexpr int MOD = 998244353;
constexpr int INF = (1LL << 62);
using minit = modint998244353;

template<typename T>
ostream& operator << (ostream& st, const vector<T> &v) {
	for (const T value : v) {
		st << value << " ";
	}
	return st;
}
template<typename T>
istream& operator >> (istream& st, vector<T>& v) {
	for (T &value : v) {
		st >> value;
	}
	return st;
}

//

//

void solve() {
	//#1 入力
	int N;
	cin >> N;
	vector<vector<int>> G(N);
	for (int i = 0; i < N - 1; i++) {
		int u, v;
		cin >> u >> v;
		u--; v--;
		G[u].push_back(v);
		G[v].push_back(u);
	}
	vector<int> color(N);
	cin >> color;

	//#2 再帰
	function<pair<int, int>(int, int)> dfs = [&](int now, int par) {
		//nowを根とする部分木を黒くするのに必要な操作回数
		// first...根は黒にできるか
		// second...最小の操作回数
		int res = 0, cnt = 0;
		for (const int next : G[now]) {
			if (next == par)continue;
			pair<int,int> p = dfs(next, now);
			cnt += 1 - p.first;
			res += p.second;
		}
		return make_pair((cnt ^ color[now]) & 1, res + cnt);
	};

	//#3 出力
	pair<int, int> p = dfs(0, -1);
	if (p.first)cout << p.second << endl;
	else cout << -1 << endl;
}
signed main() {
	int T = 1;
	// cin >> T;
	for (int i = 0; i < T; i++)
		solve();
	return 0;
}