[CodeForces - 197E] E - Paint Tree

E - Paint Tree

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 u_i and v_i (1 ≤ u_i, v_i ≤ n, u_i ≠ v_i) — the numbers of tree vertexes connected by the i-th edge.

Each of the next n lines contain two space-separated integers x_i and y_i ( - 10⁹ ≤ x_i, y_i ≤ 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.

Example

Input

31 32 30 01 12 0

Output

1 3 2

Input

41 22 31 4-1 -23 5-3 32 0

Output

4 2 1 3

Note

The possible solutions for the sample are given below.

 

题目意思就是,给你一棵已经定型的无根树,然后在二维屏幕上给出几个坐标,要求将树的每一个节点对应到二维平面上的每一个点上,要求形成的是一棵正常的树.

这题完全没思路,只知道肯定是排序+DFS,然而,就是没有想的更深一点——极角排序+分治.

我们每次递归分治的的时候,标记的总是一段区间,这个区间是有一定顺序的,他按照基准点极角排序.为什么极角排序就一定对呢?

因为每一次,你找的相当于在凸多边形上这个点的下几个点,不会穿越其他线段,这段区间也会被覆盖满(极角排序的性质很适用).最后输出就好了,效率应该是n^nlogn的。

 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #define LL long long
 using namespace std;
 ,maxe=;
 int size[maxn],fa[maxn];
 int n,tot,nxt[maxe],son[maxe],lnk[maxn],ans[maxn];
 bool vis[maxn];
 struct point{int x,y,id;}a[maxn],mainp;
 point operator - (point p,point q){point ret; ret.x=p.x-q.x,ret.y=p.y-q.y; return ret;}
 LL cross(point p,point q){return (LL)p.x*q.y-(LL)q.x*p.y;}
 bool cmp1(point u,point v){return u.x!=v.x?u.x<v.x:u.y<v.y;}
 ;}
 int read(){
     ,f=; char ch=getchar();
     '){if (ch=='-') f=-f; ch=getchar();}
     +ch-',ch=getchar();
     return x*f;
 }
 void add(int x,int y){nxt[++tot]=lnk[x],son[tot]=y,lnk[x]=tot;}
 void DFS(int x){
     vis[x]=,size[x]=;
     for (int j=lnk[x]; j; j=nxt[j]) if (!vis[son[j]]){
         DFS(son[j]); size[x]+=size[son[j]];
     }
 }
 void DFS(int x,int fa,int L,int R){
     sort(a+L,a++R,cmp1);
     ans[a[L].id]=x,mainp.x=a[L].x,mainp.y=a[L].y;
     sort(a++L,a++R,cmp2);
     int now=L;
     for (int j=lnk[x]; j; j=nxt[j]) if (son[j]!=fa){
         DFS(son[j],x,now+,now+size[son[j]]),now+=size[son[j]];
     }
 }
 int main(){
     n=read(),tot=;
     ; i<n; i++){int x=read(),y=read(); add(x,y),add(y,x);}
     memset(vis,,,);
     ; i<=n; i++) a[i].x=read(),a[i].y=read(),a[i].id=i;
     DFS(,,,n);
     ; i<=n; i++) printf("%d ",ans[i]);
     ;
 }