#pragma GCC optimize(3, "Ofast", "inline", "unroll-loops")
#include <bits/stdc++.h> 
#define int long long
using namespace std;

const int N = 300010;
const int inf = 1e18;
const int mod = 1e9 + 7;

int a[N], b[N], res[N];
struct Line
{
    int k, b;
} lne[N];
inline int calc(int x, int id)
{
    return lne[id].k * x + lne[id].b;
}
namespace My_Hyper_Sgt
{
    int indx = 0, tree[N << 2], ls[N << 2], rs[N << 2];
    inline void ins(int &rt, int l, int r, int id)
    {
        if (!rt)
        {
            tree[rt = ++indx] = id;
            return;
        }
        int mid = l + r >> 1;
        if (calc(mid, tree[rt]) > calc(mid, id))
            swap(tree[rt], id);
        if (calc(l, tree[rt]) > calc(l, id))
            ins(ls[rt], l, mid, id);
        if (calc(r, tree[rt]) > calc(r, id))
            ins(rs[rt], mid + 1, r, id);
    }
    inline int qry(int rt, int l, int r, int x)
    {
        if (!rt)
            return inf;
        int mid = l + r >> 1;
        if (x <= mid)
            return min(calc(x, tree[rt]), qry(ls[rt], l, mid, x));
        return min(calc(x, tree[rt]), qry(rs[rt], mid + 1, r, x));
    }
} using namespace My_Hyper_Sgt;
vector<int> adj[N];
inline int merge(int x, int y, int l, int r)
{
    if (!x || !y)
        return x | y;
    ins(x, l, r, tree[y]);
    int mid = l + r >> 1;
    ls[x] = merge(ls[x], ls[y], l, mid);
    rs[x] = merge(rs[x], rs[y], mid + 1, r);
    return x;
}
inline void dfs(int u, int fa)
{
    for (int &v : adj[u])
        if (v != fa)
        {
            dfs(v, u);
            tree[u] = merge(tree[u], tree[v], 1, N - 10);
        }
    res[u] = qry(tree[u], 1, N - 10, a[u] + 100010);
    // cout << "res[" << u << "] = " << res[u] << '\n';
    if (res[u] >= inf)
        res[u] = 0;
    lne[u] = {b[u], res[u] - b[u] * 100010};
    ins(tree[u], 1, N - 10, u);
}
signed main()
{
    // freopen("1.in", "r", stdin);
    // freopen("1.out", "w", stdout);
    cin.tie(0)->sync_with_stdio(false);
    cout << fixed << setprecision(15);
    int n;
    cin >> n;
    for (int i = 1; i <= n; ++i)
        cin >> a[i];
    for (int i = 1; i <= n; ++i)
        cin >> b[i];
    for (int i = 1; i < n; ++i)
    {
        int a, b;
        cin >> a >> b;
        adj[a].emplace_back(b);
        adj[b].emplace_back(a);
    }
    dfs(1, 0);
    for (int i = 1; i <= n; ++i)
        cout << res[i] << ' ';
    cout << '\n';
    return 0;
}

求调/kel