#include <cstdio>
#include <vector>
#define N 100001
using namespace std;
int n, k[N], d[N];
vector<int> a[N];
vector<int> b;
bool use[N];
void read(){
	scanf("%d ", &n);
	for(int i = 1; i <= n; ++i)
		scanf("%d ", &k[i]);
	int x, y;
	for(int i = 1; i < n; ++i){
		scanf("%d %d ", &x, &y);
		use[y] = true;
		a[x].push_back(y);
	}
}
bool used[N];
void dfs(int t){
	used[t] = true;
	b.push_back(t);
	d[t] = d[b[b.size() - k[t] - 1]] + 1;
	for(vector<int>::iterator it = a[t].begin(); it != a[t].end(); ++it)
		if(!used[*it])
			dfs(*it);
	b.pop_back();
}
void solve(){
	for(int i = 1; i <= n; ++i)
		if(!use[i]){
			dfs(i);
			break;
		}
	for(int i = 1; i <= n; ++i)
		printf("%d ", d[i] - 1);
	printf("\n");
}
int main(){
	freopen("cerere.in", "r", stdin);
freopen("cerere.out", "w", stdout);
	read();
	solve();
	return 0;
}
//tot am niste raspunsuri gresite o.O 
// nu calculam bine radacina, nu e aia cu k[i] == 0 ca radacina e unica.
// el iti da muchii orientate i -> j (se numesc arce)
// radacina e acel i carea are grad intern 0, adica nu vine nicio muchie in el
// deci tre sa vezi care nod nu primeste niciodata un arc... adica sa nu apara in postura de y la citire :da)