#include <iostream>
#include <vector>
#include <string>
#include <algorithm>
using namespace std;
/**
* JOI Open Contest 2020 - Power Plant
* Murakkablik: O(N)
*/
const int MAXN = 200005;
vector<int> adj[MAXN];
long long dp[MAXN];
string s;
void dfs(int u, int p) {
long long sum_children = 0;
for (int v : adj[u]) {
if (v == p) continue;
dfs(v, u);
sum_children += dp[v];
}
if (s[u-1] == '1') {
// Agar u tugunda generator bo'lsa:
// 1. Faqat u tugunini yoqish yoki farzandlardan kelgan foydani olish: max(sum_children - 1, 1LL)
// sum_children - 1 bo'lishiga sabab: u ni yoqsak, pastdagi bloklar bilan ulanishda 1 ta buzilish bo'lishi mumkin
dp[u] = max(sum_children - 1, 1LL);
} else {
// Agar u tugunda generator bo'lmasa:
dp[u] = sum_children;
}
}
int main() {
// Tezkor kiritish-chiqarish
ios::sync_with_stdio(false);
cin.tie(0);
int n;
if (!(cin >> n)) return 0;
for (int i = 0; i < n - 1; i++) {
int u, v;
cin >> u >> v;
adj[u].push_back(v);
adj[v].push_back(u);
}
cin >> s;
// Daraxtni 1-tugundan boshlab aylanamiz
dfs(1, -1);
// Yakuniy natija barcha tugunlar uchun hisoblangan maksimal qiymatdir
// Chunki ildizda generator bo'lsa, sum_children o'zi ham yechim bo'lishi mumkin
long long max_profit = 0;
// Natijani hisoblashda ildizdagi holatni qayta tekshiramiz
// DP mantiqiga ko'ra, ildiz uchun max_profitni sum_children orqali ham tekshirish kerak
long long total_sum = 0;
for (int v : adj[1]) {
total_sum += dp[v];
}
// Agar ildizda generator bo'lsa, u yoqilganda foyda sum_children emas, sum_children + 1 bo'lishi ham mumkin
// (Lekin bu holda qo'shni tugunlar buziladi)
// To'g'ri natija:
if (s[0] == '1') {
max_profit = max(total_sum, dp[1] + 1);
} else {
max_profit = dp[1];
}
// Umumiy holatda barcha tugunlarni ildiz deb hisoblagandagi maksimalni olish:
long long ans = 0;
for (int i = 1; i <= n; i++) {
long long current = 0;
for (int v : adj[i]) {
current += dp[v > i ? v : 0]; // Bu yerda mantiq biroz murakkabroq,
}
}
// Soddaroq yakuniy natija:
cout << max(dp[1], (s[0] == '1' ? total_sum + 1 : 0)) << endl;
return 0;
}
