#include <bits/stdc++.h>
using namespace std;
int n, s[1510];
vector<pair<int, pair<int, int> > > x;
vector<int> e[1510];
int c[1510], par[1510];
int curnode;
double cval[1510];
bool cmp(pair<int, pair<int, int> > a, pair<int, pair<int, int> > b) {
  return cval[a.first] < cval[b.first];
}
bool cY(pair<int, pair<int, int> > a, pair<int, pair<int, int> > b) {
  return a.second.second < b.second.second;
}
void dp(int treenode, vector<pair<int, pair<int, int> > > &nodes) {
  int cmin = 0;
  for (int i = 0; i < nodes.size(); i++) {
    if (nodes[i].second.second < nodes[cmin].second.second) cmin = i;
  }
  swap(nodes[cmin], nodes[0]);
  curnode = nodes[0].first;
  c[curnode] = treenode;
  double pivotX = nodes[0].second.first;
  double pivotY = nodes[0].second.second;
  nodes.erase(nodes.begin());
  for (int i = 0; i < nodes.size(); i++) {
    pair<int, pair<int, int> > j = nodes[i];
    cval[j.first] = atan2(j.second.second - pivotY, j.second.first - pivotX);
  }
  sort(nodes.begin(), nodes.end(), cmp);
  for (int i = 0; i < e[treenode].size(); i++) {
    int j = e[treenode][i];
    if (j != par[treenode]) {
      vector<pair<int, pair<int, int> > > nxt;
      for (int k = 0; k < s[j]; k++) {
        nxt.push_back(nodes.back());
        nodes.pop_back();
      }
      dp(j, nxt);
    }
  }
}
int dfs(int i, int p) {
  par[i] = p;
  s[i] = 1;
  for (auto &j : e[i]) {
    if (j != p) s[i] += dfs(j, i);
  }
  return s[i];
}
int32_t main() {
  cin >> n;
  for (int i = 1; i < n; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    e[a].push_back(b);
    e[b].push_back(a);
  }
  x.resize(n);
  for (int i = 0; i < n; i++) {
    x[i].first = i;
    cin >> x[i].second.first >> x[i].second.second;
  }
  sort(x.begin(), x.end(), cY);
  dfs(0, 0);
  dp(0, x);
  for (int i = 0; i < n; i++) cout << c[i] + 1 << " ";
  return 0;
}

### Prompt

Generate a Cpp solution to the following problem:

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;
int n, s[1510];
vector<pair<int, pair<int, int> > > x;
vector<int> e[1510];
int c[1510], par[1510];
int curnode;
double cval[1510];
bool cmp(pair<int, pair<int, int> > a, pair<int, pair<int, int> > b) {
  return cval[a.first] < cval[b.first];
}
bool cY(pair<int, pair<int, int> > a, pair<int, pair<int, int> > b) {
  return a.second.second < b.second.second;
}
void dp(int treenode, vector<pair<int, pair<int, int> > > &nodes) {
  int cmin = 0;
  for (int i = 0; i < nodes.size(); i++) {
    if (nodes[i].second.second < nodes[cmin].second.second) cmin = i;
  }
  swap(nodes[cmin], nodes[0]);
  curnode = nodes[0].first;
  c[curnode] = treenode;
  double pivotX = nodes[0].second.first;
  double pivotY = nodes[0].second.second;
  nodes.erase(nodes.begin());
  for (int i = 0; i < nodes.size(); i++) {
    pair<int, pair<int, int> > j = nodes[i];
    cval[j.first] = atan2(j.second.second - pivotY, j.second.first - pivotX);
  }
  sort(nodes.begin(), nodes.end(), cmp);
  for (int i = 0; i < e[treenode].size(); i++) {
    int j = e[treenode][i];
    if (j != par[treenode]) {
      vector<pair<int, pair<int, int> > > nxt;
      for (int k = 0; k < s[j]; k++) {
        nxt.push_back(nodes.back());
        nodes.pop_back();
      }
      dp(j, nxt);
    }
  }
}
int dfs(int i, int p) {
  par[i] = p;
  s[i] = 1;
  for (auto &j : e[i]) {
    if (j != p) s[i] += dfs(j, i);
  }
  return s[i];
}
int32_t main() {
  cin >> n;
  for (int i = 1; i < n; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    e[a].push_back(b);
    e[b].push_back(a);
  }
  x.resize(n);
  for (int i = 0; i < n; i++) {
    x[i].first = i;
    cin >> x[i].second.first >> x[i].second.second;
  }
  sort(x.begin(), x.end(), cY);
  dfs(0, 0);
  dp(0, x);
  for (int i = 0; i < n; i++) cout << c[i] + 1 << " ";
  return 0;
}

```