// CEOI 2022 Day2 Drawing
#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
struct Point {
int x, y, id;
bool operator<(const Point& p) const {
if (x != p.x) return x < p.x;
return y < p.y;
}
};
LL cross(const Point& a, const Point& b, const Point& c) { // cross of a→b, a→c
LL x1 = b.x - a.x, y1 = b.y - a.y, x2 = c.x - a.x, y2 = c.y - a.y;
return x1 * y2 - x2 * y1;
}
int main() {
ios::sync_with_stdio(false), cin.tie(0);
int N;
cin >> N;
vector<vector<int>> G(N + 1);
for (int i = 1, a, b; i < N; i++)
cin >> a >> b, G[a].push_back(b), G[b].push_back(a);
vector<Point> ps(N);
for (int i = 0; i < N; i++) cin >> ps[i].x >> ps[i].y, ps[i].id = i + 1;
sort(begin(ps), end(ps)); // O=ps[0] 变成最左下方的点O
auto pi = begin(ps);
nth_element(pi, pi + 1, end(ps));
sort(pi + 1, end(ps), // 以O为原点, 所有按照极坐标逆时针排序
[&](const Point& a, const Point& b) { return cross(ps[0], a, b) >= 0; });
vector<int> Ans(N + 1);
function<void(int, int)> dfs = [&](int u, int fa) {
Ans[u] = pi->id, pi++;
for (int v : G[u])
if (v != fa) dfs(v, u);
};
dfs(1, 0);
for (int i = 1; i <= N; i++) printf("%d ", Ans[i]);
puts("");
return 0;
}
// AC 100
