#include <bits/stdc++.h>
using namespace std;
int n, res;
const int MAXN = 200'050;
vector<int> v[MAXN];
string s;
bool has_gen[MAXN];
int dfs(int x, int p) {
int mx = 0, sum = 0;
for (auto z : v[x]) {
if (z != p) {
int dp = dfs(z, x);
mx = max(mx, dp);
sum += dp;
}
}
int ans;
// My key concern was that if we assume there is
// an ON node above x, but there actually isn't.
// But in ans we can pretend x is the root node
// E.g. if there is actually no ON node above x
// then one possible case for ans would be
// mx + 1 (ON x if has_gen[x]) (although dp is still 1)
// Basically also calculate ans at each node assuming the root of the tree is x besides dp
if (!has_gen[x]) {
ans = sum;
res = max(res, ans);
return sum; // = dp
}
// Force generator ON
ans = 1 + mx; // Assuming x is the root (i.e. no ON above x)
int ret = 1; // Assuming has ON above x, we must pick only 1
// Force generator OFF
ans = max(ans, sum - 1); // Again assuming x is the root
ret = max(ret, sum - 1); // Even if there is ON above x it's still this. Here we maximise dp before returning it
res = max(res, ans);
return ret;
}
int main() {
ios::sync_with_stdio(false);
cin.tie(0);
cin >> n;
for (int i = 0, a, b; i < n - 1; i++) {
cin >> a >> b;
v[a].push_back(b);
v[b].push_back(a);
}
cin >> s;
for (int i = 1; i <= n; i++) {
has_gen[i] = s[i - 1] == '1';
}
dfs(1, -1);
cout << res;
return 0;
}
