#include<bits/stdc++.h>
using namespace std;
#define int long long
#define fi first
#define se second
#define pb push_back
#define mp make_pair
#define foru(i, l, r) for(int i = l; i <= r; i++)
#define ford(i, r, l) for(int i = r; i >= l; i--)
typedef pair<int, int> ii;
typedef pair<ii, int> iii;
typedef pair<ii, ii> iiii;
const int N = 2e5 + 5;
const int oo = 1e18 + 7, mod = 1e9 + 7;
int n;
bool plant[N];
vector<int> Adj[N];
/*
Basically, there are three main cases:
- Not choosing subtree of u (including u)
- Just choosing subtree of u
- Not just choosing subtree of u -> every node that is parent of at least one node in subtree u is broken
*/
int dp[N][2];// cases above (ofc first case -> 0)
int par[N];
void dfs(int u, int p){
par[u] = p;
for(auto v : Adj[u]){
if(v == p) continue;
dfs(v, u);
}
int mx = 0;
for(auto v : Adj[u]){
if(v == p) continue;
dp[u][0] += max(dp[v][1], 0LL);// you can not choose anything in v
mx = max(mx, dp[v][1]);
}
dp[u][0] -= plant[u];
dp[u][0] = max(dp[u][0], mx + plant[u]);// just choosing one subtree of u => you can choose u
dp[u][1] = plant[u];// just choosing u
int sum = 0;
for(auto v : Adj[u]){
if(v == p) continue;
sum += max(0LL, dp[v][1]);
}
sum -= plant[u];
dp[u][1] = max(dp[u][1], sum);
}
void process(){
cin >> n;
for(int i = 1; i < n; i++){
int x, y;
cin >> x >> y;
Adj[x].pb(y);
Adj[y].pb(x);
}
string s;
cin >> s;
for(int i = 1; i <= n; i++) plant[i] = s[i - 1] - '0';
dfs(1, 1);
int mx = -1;
for(int i = 1; i <= n; i++) mx = max(mx, dp[i][0]);
cout << mx;
}
signed main(){
ios_base::sync_with_stdio(0);
process();
}
