#include <bits/stdc++.h>
using namespace std;
int tot = 0, n, root;
int x[1505 * 2], y[1505 * 2], pre[1505 * 2], last[1505], size[1505], ans[1505];
int c1[1505], c2[1505], b[1505], s;
void add(int u, int v) {
  x[++tot] = u;
  y[tot] = v;
  pre[tot] = last[u];
  last[u] = tot;
}
inline long long cross(int t1, int t2, int t3, int t4, int t5, int t6) {
  return (long long)(t3 - t1) * (long long)(t6 - t2) -
         (long long)(t4 - t2) * (long long)(t5 - t1);
}
inline long long sqr(int v) { return (long long)v * v; }
inline long long dis(int t1, int t2, int t3, int t4) {
  return sqr(t3 - t1) + sqr(t4 - t2);
}
bool comp(int i, int j) {
  return cross(c1[s], c2[s], c1[i], c2[i], c1[j], c2[j]) > 0 ||
         (cross(c1[s], c2[s], c1[i], c2[i], c1[j], c2[j]) == 0 &&
          dis(c1[s], c2[s], c1[i], c2[i]) < dis(c1[s], c2[s], c1[j], c2[j]));
}
void init() {
  int i, t1, t2;
  scanf("%d", &n);
  for (i = 1; i <= n - 1; i++) {
    scanf("%d%d", &t1, &t2);
    add(t1, t2);
    add(t2, t1);
  }
  for (i = 1; i <= n; i++) b[i] = i;
  for (i = 1; i <= n; i++) scanf("%d%d", c1 + i, c2 + i);
}
void dfs(int v, int fa) {
  size[v] = 1;
  for (int w = last[v]; w; w = pre[w])
    if (y[w] != fa) {
      dfs(y[w], v);
      size[v] += size[y[w]];
    }
}
void calc(int v, int fa, int s1, int s2) {
  int i, tmp = s1, q = s1 + 1;
  for (i = s1; i <= s2; i++) {
    if (c1[b[i]] < c1[b[tmp]] ||
        (c1[b[i]] == c1[b[tmp]] && c2[b[i]] < c2[b[tmp]]))
      tmp = i;
  }
  swap(b[s1], b[tmp]);
  s = b[s1];
  sort(b + s1 + 1, b + s2 + 1, comp);
  ans[b[s1]] = v;
  for (int w = last[v]; w; w = pre[w])
    if (y[w] != fa) {
      calc(y[w], v, q, q + size[y[w]] - 1);
      q += size[y[w]];
    }
}
void work() {
  dfs(1, 0);
  calc(1, 0, 1, n);
  int i;
  for (i = 1; i <= n; i++) printf("%d ", ans[i]);
}
int main() {
  init();
  work();
}

### Prompt

Create a solution in cpp for 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 tot = 0, n, root;
int x[1505 * 2], y[1505 * 2], pre[1505 * 2], last[1505], size[1505], ans[1505];
int c1[1505], c2[1505], b[1505], s;
void add(int u, int v) {
  x[++tot] = u;
  y[tot] = v;
  pre[tot] = last[u];
  last[u] = tot;
}
inline long long cross(int t1, int t2, int t3, int t4, int t5, int t6) {
  return (long long)(t3 - t1) * (long long)(t6 - t2) -
         (long long)(t4 - t2) * (long long)(t5 - t1);
}
inline long long sqr(int v) { return (long long)v * v; }
inline long long dis(int t1, int t2, int t3, int t4) {
  return sqr(t3 - t1) + sqr(t4 - t2);
}
bool comp(int i, int j) {
  return cross(c1[s], c2[s], c1[i], c2[i], c1[j], c2[j]) > 0 ||
         (cross(c1[s], c2[s], c1[i], c2[i], c1[j], c2[j]) == 0 &&
          dis(c1[s], c2[s], c1[i], c2[i]) < dis(c1[s], c2[s], c1[j], c2[j]));
}
void init() {
  int i, t1, t2;
  scanf("%d", &n);
  for (i = 1; i <= n - 1; i++) {
    scanf("%d%d", &t1, &t2);
    add(t1, t2);
    add(t2, t1);
  }
  for (i = 1; i <= n; i++) b[i] = i;
  for (i = 1; i <= n; i++) scanf("%d%d", c1 + i, c2 + i);
}
void dfs(int v, int fa) {
  size[v] = 1;
  for (int w = last[v]; w; w = pre[w])
    if (y[w] != fa) {
      dfs(y[w], v);
      size[v] += size[y[w]];
    }
}
void calc(int v, int fa, int s1, int s2) {
  int i, tmp = s1, q = s1 + 1;
  for (i = s1; i <= s2; i++) {
    if (c1[b[i]] < c1[b[tmp]] ||
        (c1[b[i]] == c1[b[tmp]] && c2[b[i]] < c2[b[tmp]]))
      tmp = i;
  }
  swap(b[s1], b[tmp]);
  s = b[s1];
  sort(b + s1 + 1, b + s2 + 1, comp);
  ans[b[s1]] = v;
  for (int w = last[v]; w; w = pre[w])
    if (y[w] != fa) {
      calc(y[w], v, q, q + size[y[w]] - 1);
      q += size[y[w]];
    }
}
void work() {
  dfs(1, 0);
  calc(1, 0, 1, n);
  int i;
  for (i = 1; i <= n; i++) printf("%d ", ans[i]);
}
int main() {
  init();
  work();
}

```