#include <bits/stdc++.h>
using namespace std;
struct point {
  int x, y, id;
  bool operator<(point other) const {
    return ((x < other.x) || (x == other.x && y < other.y));
  }
};
const int N = 1505;
const int inf = (int)2e9 + 10;
int n;
vector<int> adj[N];
point a[N];
int nChild[N], res[N], dad[N];
void DFS(int u, int father) {
  int v;
  dad[u] = father;
  nChild[u] = 1;
  for (int i = 0; i < adj[u].size(); i++) {
    v = adj[u][i];
    if (v == father) continue;
    DFS(v, u);
    nChild[u] += nChild[v];
  }
}
bool cmp(point a, point b) {
  return (long long)a.y * b.x > (long long)b.y * a.x;
}
void Build(int l, int r, int u) {
  sort(a + l, a + r + 1);
  res[a[l].id] = u;
  for (int i = l + 1; i <= r; i++) {
    a[i].x -= a[l].x;
    a[i].y -= a[l].y;
  }
  sort(a + l + 1, a + r + 1, cmp);
  int v, last = 1;
  for (int i = 0; i < adj[u].size(); i++) {
    v = adj[u][i];
    if (v == dad[u]) continue;
    Build(l + last, l + last + nChild[v] - 1, v);
    last += nChild[v];
  }
}
int main() {
  cin >> n;
  int u, v;
  for (int i = 1; i < n; i++) {
    scanf("%d %d", &u, &v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d %d", &a[i].x, &a[i].y);
    a[i].id = i;
  }
  DFS(1, 0);
  Build(1, n, 1);
  for (int i = 1; i <= n; i++) printf("%d ", res[i]);
  return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

You are given a tree with n vertexes and n points on a plane, no three points lie on one straight line.

Your task is to paint the given tree on a plane, using the given points as vertexes. 

That is, you should correspond each vertex of the tree to exactly one point and each point should correspond to a vertex. If two vertexes of the tree are connected by an edge, then the corresponding points should have a segment painted between them. The segments that correspond to non-adjacent edges, should not have common points. The segments that correspond to adjacent edges should have exactly one common point.

Input

The first line contains an integer n (1 ≤ n ≤ 1500) — the number of vertexes on a tree (as well as the number of chosen points on the plane).

Each of the next n - 1 lines contains two space-separated integers ui and vi (1 ≤ ui, vi ≤ n, ui ≠ vi) — the numbers of tree vertexes connected by the i-th edge.

Each of the next n lines contain two space-separated integers xi and yi ( - 109 ≤ xi, yi ≤ 109) — the coordinates of the i-th point on the plane. No three points lie on one straight line.

It is guaranteed that under given constraints problem has a solution.

Output

Print n distinct space-separated integers from 1 to n: the i-th number must equal the number of the vertex to place at the i-th point (the points are numbered in the order, in which they are listed in the input).

If there are several solutions, print any of them.

Examples

Input

3
1 3
2 3
0 0
1 1
2 0


Output

1 3 2


Input

4
1 2
2 3
1 4
-1 -2
3 5
-3 3
2 0


Output

4 2 1 3

Note

The possible solutions for the sample are given below.

<image> <image>

### Response

```cpp
#include <bits/stdc++.h>
using namespace std;
struct point {
  int x, y, id;
  bool operator<(point other) const {
    return ((x < other.x) || (x == other.x && y < other.y));
  }
};
const int N = 1505;
const int inf = (int)2e9 + 10;
int n;
vector<int> adj[N];
point a[N];
int nChild[N], res[N], dad[N];
void DFS(int u, int father) {
  int v;
  dad[u] = father;
  nChild[u] = 1;
  for (int i = 0; i < adj[u].size(); i++) {
    v = adj[u][i];
    if (v == father) continue;
    DFS(v, u);
    nChild[u] += nChild[v];
  }
}
bool cmp(point a, point b) {
  return (long long)a.y * b.x > (long long)b.y * a.x;
}
void Build(int l, int r, int u) {
  sort(a + l, a + r + 1);
  res[a[l].id] = u;
  for (int i = l + 1; i <= r; i++) {
    a[i].x -= a[l].x;
    a[i].y -= a[l].y;
  }
  sort(a + l + 1, a + r + 1, cmp);
  int v, last = 1;
  for (int i = 0; i < adj[u].size(); i++) {
    v = adj[u][i];
    if (v == dad[u]) continue;
    Build(l + last, l + last + nChild[v] - 1, v);
    last += nChild[v];
  }
}
int main() {
  cin >> n;
  int u, v;
  for (int i = 1; i < n; i++) {
    scanf("%d %d", &u, &v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  for (int i = 1; i <= n; i++) {
    scanf("%d %d", &a[i].x, &a[i].y);
    a[i].id = i;
  }
  DFS(1, 0);
  Build(1, n, 1);
  for (int i = 1; i <= n; i++) printf("%d ", res[i]);
  return 0;
}

```