#include <vector>
#include <algorithm>
#include <stdio.h>
#include <ctype.h>
#include <cstring>
#include <climits>

#pragma GCC optimize ("O1")
#pragma GCC optimize ("O2")
#pragma GCC optimize ("O3")
#pragma GCC optimize ("unroll-loops")
#pragma GCC optimize ("Ofast")

using namespace std;

class InParser
{
private:
    FILE *fin;
    char *buff;
    int sp;
    char read_ch()
    {
        ++sp;
        if (sp == 4096)
        {
            sp = 0;
            fread(buff, 1, 4096, fin);
        }
        return buff[sp];
    }
public:
    InParser(const char* nume)
    {
        fin = fopen(nume, "r");
        buff = new char[4096]();
        sp = 4095;
    }
    InParser& operator >> (int &n)
    {
        char c;
        while (!isdigit(c = read_ch()) && c != '-');
        int sgn = 1;
        if (c == '-')
        {
            n = 0;
            sgn = -1;
        }
        else
        {
            n = c - '0';
        }
        while (isdigit(c = read_ch()))
        {
            n = 10 * n + c - '0';
        }
        n *= sgn;
        return *this;
    }
    InParser& operator >> (long long &n)
    {
        char c;
        n = 0;
        while (!isdigit(c = read_ch()) && c != '-');
        long long sgn = 1;
        if (c == '-')
        {
            n = 0;
            sgn = -1;
        }
        else
        {
            n = c - '0';
        }
        while (isdigit(c = read_ch()))
        {
            n = 10 * n + c - '0';
        }
        n *= sgn;
        return *this;
    }
};

class OutParser
{
private:
    FILE *fout;
    char *buff;
    int sp;
    void write_ch(char ch)
    {
        if (sp == 50000)
        {
            fwrite(buff, 1, 50000, fout);
            sp = 0;
            buff[sp++] = ch;
        }
        else
        {
            buff[sp++] = ch;
        }
    }
public:
    OutParser(const char* name)
    {
        fout = fopen(name, "w");
        buff = new char[50000]();
        sp = 0;
    }
    ~OutParser()
    {
        fwrite(buff, 1, sp, fout);
        fclose(fout);
    }
    OutParser& operator << (int vu32)
    {
        if (vu32 <= 9)
        {
            write_ch(vu32 + '0');
        }
        else
        {
            (*this) << (vu32 / 10);
            write_ch(vu32 % 10 + '0');
        }
        return *this;
    }
    OutParser& operator << (long long vu64)
    {
        if (vu64 <= 9)
        {
            write_ch(vu64 + '0');
        }
        else
        {
            (*this) << (vu64 / 10);
            write_ch(vu64 % 10 + '0');
        }
        return *this;
    }
    OutParser& operator << (char ch)
    {
        write_ch(ch);
        return *this;
    }
    OutParser& operator << (const char *ch)
    {
        while (*ch)
        {
            write_ch(*ch);
            ++ch;
        }
        return *this;
    }
};

InParser in ("cerere.in");
OutParser out ("cerere.out");

const int max_size = 1e5 + 1, max_r = 18;

int t[max_r][max_size], a[max_size], ans[max_size];
bool tata[max_size];
/// cel care ramane 0 e radacina => pt dfs si nivelare
vector <int> edges[max_size];
vector <pair <int, int>> lvl;

void dfs (int nod, int nivel)
{
    lvl.push_back({nivel, nod});
    for (auto f : edges[nod])
    {
        dfs(f, nivel + 1);
    }
}

int anc (int x, int ord)
{
    int e = 0;
    while (ord > 0)
    {
        if (ord % 2 == 1)
        {
            x = t[e][x];
        }
        e++;
        ord /= 2;
    }
    return x;
}

int main ()
{
    int n;
    in >> n;
    for (int i = 1; i <= n; i++)
    {
        in >> a[i];
    }
    for (int i = 1; i < n; i++)
    {
        int x, y;
        in >> x >> y;
        edges[x].push_back(y);
        t[0][y] = x;
        tata[y] = true;
    }
    for (int e = 1; e < max_r; e++)
    {
        for (int i = 1; i <= n; i++)
        {
            t[e][i] = t[e - 1][t[e - 1][i]];
        }
    }
    for (int i = 1; i <= n; i++)
    {
        if (!tata[i])
        {
            dfs(i, 1);
            break;
        }
    }
    sort(lvl.begin(), lvl.end());
    for (int i = 1; i < n; i++)
    {
        if (a[lvl[i].second] > 0)
        {
            ans[lvl[i].second] = 1 + ans[anc(lvl[i].second, a[lvl[i].second])];
        }
    }
    for (int i = 1; i <= n; i++)
    {
        out << ans[i] << " ";
    }
    return 0;
}