#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);
// Faqat ijobiy foyda keltiradigan ostki daraxtlarni qo'shamiz
if (dp[v] > 0) {
sum_children += dp[v];
}
}
if (s[u-1] == '1') {
// Agar tugunda generator bo'lsa:
// sum_children - 1: u ni yoqsak, pastdagi bloklar bilan ulanishda 1 ta ta'mir xarajatlari chiqadi [cite: 19, 64]
// 1LL: Faqat shu tugunni yoqib, pastdagilarni o'chirish [cite: 17]
dp[u] = max(sum_children - 1, 1LL);
} else {
// Agar generator bo'lmasa, shunchaki farzandlar foydasi [cite: 20]
dp[u] = sum_children;
}
}
int main() {
// Kiritish va chiqarishni tezlashtirish
ios::sync_with_stdio(false);
cin.tie(NULL);
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 hisoblaymiz
dfs(1, -1);
// Ildiz tugunida maxsus holatni tekshiramiz:
// Agar ildizda generator bo'lsa, uni yoqish orqali barcha farzandlardan
// kelgan ijobiy foydani buzilishsiz olish mumkin (sum_children + 1)
long long root_sum = 0;
for (int v : adj[1]) {
if (dp[v] > 0) root_sum += dp[v];
}
long long ans = dp[1];
if (s[0] == '1') {
ans = max(ans, root_sum + 1);
}
cout << ans << endl;
return 0;
}
