#include <bits/stdc++.h>

using namespace std;
typedef long long ll;
typedef unsigned long long ull;

#define N 200010
int n;
int w[N];
int p[N];
vector<int>e[N];
int rt[N];
int ls[N << 5], rs[N << 5], sz[N << 5];
int val[N << 5], cnt;

#define mid ((l + r) >> 1)

void add(int &x, int l, int r, int pos)
{
    if (!x) {x = ++ cnt; sz[x] = 1;}
    if (l == r) return ;
    if (pos <= mid) add(ls[x], l, mid, pos);
    else add(rs[x], mid + 1, r, pos);
}
void pushdown(int x)
{
    if (val[x])
    {
        if (ls[x])val[ls[x]] += val[x];
        if (rs[x])val[rs[x]] += val[x];
        val[x] = 0;
    }
}
int mer(int x, int y, int l, int r, int szx, int szy, int w)
{
    // cout << "EE:" << x << ' ' << y << ' ' << szx << ' ' << szy << ' ' << l << ' ' << r << ' ' << w << endl;
    if (x == 0 && y == 0) return 0;
    else if (x == 0)
    {
        val[y] += w * szx;
        return y;
    }
    else if (y == 0)
    {
        val[x] += w * szy;
        return x;
    }
    pushdown(x); pushdown(y);
    rs[x] = mer(rs[x], rs[y], mid + 1, r, szx + sz[ls[x]], szy + sz[ls[y]], w);
    ls[x] = mer(ls[x], ls[y], l, mid, szx, szy, w);

    sz[x] = sz[ls[x]] + sz[rs[x]];
    return x;
}
void dfs(int u)
{
    add(rt[u], 1, n, u);
    for (int v : e[u]) 
    {
        dfs(v);
        // cout << "MERGE:" << u << " " << v << endl;
        rt[u] = mer(rt[u], rt[v], 1, n, 0, 0, w[u]);
    }
}
void sol(int x, int l, int r, int s)
{
    // cout << "???:" << x << ' ' << l << ' ' << r << ' ' << s << endl;
    s += val[x];
    if (l == r)
    {
        if (l > 1)cout << s << '\n';
        return ;
    }
    
    sol(ls[x], l, mid, s); sol(rs[x], mid + 1, r, s);
}
void sol()
{
    for (int i = 1; i <= n; i ++) cin >> w[i];
    for (int i = 2; i <= n; i ++) 
    {
        cin >> p[i];
        e[p[i]].push_back(i);
    }
    dfs(1);
    sol(rt[1], 1, n, 0);

    for (int i = 1; i <= n; i ++)e[i].clear(), rt[i] = 0;
    for (int i = 1; i <= cnt; i ++)sz[i] = ls[i] = rs[i] = val[i] = 0;
    cnt = 0;
} 

int main()
{
    ios::sync_with_stdio(0); cin.tie(0);
    // int T = 1;
    // cin >> T;
    while (cin >> n) sol();
    return 0;
}