/*
* Observe that you can only have one node be a parent of another. Let dp[v] = max profit you can get if no chosen node is a parent of another chosen node.
*
* Then, if node v has a generator, dp[v] = sum of dp[child_v] - 1, and if it doesn't, dp[v] = sum of dp[child_v].
*
* Finally, since you can have one node be the parent of another, ans = max(ans, dp[v], dp[child_v] + 1) over all v.
*/
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using vi = vector<int>;
#define pb push_back
#define rsz resize
#define all(x) begin(x), end(x)
#define sz(x) (int)(x).size()
using pi = pair<int,int>;
#define f first
#define s second
#define mp make_pair
const int MX = 200005;
vi adj[MX];
string s;
ll ans = 0;
ll dp[MX];
void dfs(int v, int p) {
if (s[v] == '1') {
ll best = 0;
for (int to : adj[v]) {
if (to != p) {
dfs(to, v);
dp[v] += dp[to];
best = max(best, dp[to] + 1);
}
}
ans = max(ans, best);
dp[v]--;
dp[v] = max(dp[v], 1LL);
}
else {
for (int to : adj[v]) {
if (to != p) {
dfs(to, v);
dp[v] += dp[to];
}
}
}
ans = max(ans, dp[v]);
}
int main(){
ios::sync_with_stdio(false);
cin.tie(0);
int n; cin >> n;
for (int i = 0; i < n; i++) {
int u, v; cin >> u >> v;
u--, v--;
adj[u].pb(v), adj[v].pb(u);
}
cin >> s;
dfs(0, -1);
cout << ans << '\n';
}
