CF1009F Dominant Indices（启发式合并）

You are given a rooted undirected tree consisting of nn vertices. Vertex 11 is the root.

Let's denote a depth array of vertex xx as an infinite sequence [dx,0,dx,1,dx,2,…][dx,0,dx,1,dx,2,…], where dx,idx,i is the number of vertices yy such that both conditions hold:

• xx is an ancestor of yy;
• the simple path from xx to yy traverses exactly ii edges.

The dominant index of a depth array of vertex xx (or, shortly, the dominant index of vertex xx) is an index jj such that:

• for every k<jk<j, dx,k<dx,jdx,k<dx,j;
• for every k>jk>j, dx,k≤dx,jdx,k≤dx,j.

For every vertex in the tree calculate its dominant index.

Input

The first line contains one integer nn (1≤n≤1061≤n≤106) — the number of vertices in a tree.

Then n−1n−1 lines follow, each containing two integers xx and yy (1≤x,y≤n1≤x,y≤n, x≠yx≠y). This line denotes an edge of the tree.

It is guaranteed that these edges form a tree.

Output

Output nn numbers. ii-th number should be equal to the dominant index of vertex ii.

Examples

Input

4
1 2
2 3
3 4

Output

0
0
0
0

Input

4
1 2
1 3
1 4

Output

1
0
0
0

Input

4
1 2
2 3
2 4

Output

2
1
0
0
题解：题目意思为：给你一颗有根数，对于每一个节点，他的子树到他的边数为d,到达他的边数为d的数量为fd,让你求每一个节点fd最大的d,如果有多个选择d最小的那个；
用map<int,int> mp[i][d]：表示节点I的子树中的节点中深度为d（表示到根节点的边数）的数量。然后沿着重边一直往下搜索，启发式搜索，每次更新ans[u]即可；
参考代码：

 #include<bits/stdc++.h>
 using namespace std;
 #define clr(a,val) memset(a,val,sizeof(a))
 #define pb push_back
 #define fi first
 #define se second
 typedef long long ll;
 const int maxn=1e6+;
 inline int read()
 {
     int x=,f=;char ch=getchar();
     while(ch<''||ch>''){if(ch=='-')f=-;ch=getchar();}
     while(ch>=''&&ch<=''){x=(x<<)+(x<<)+ch-'';ch=getchar();}
     return x*f;
 }
 struct Edge{
     int to,nxt;
 } edge[maxn<<];
 int n,cnt,dep[maxn],hvy[maxn],hson[maxn],head[maxn<<];
 int maxd[maxn],ans[maxn];
 map<int,int> mp[maxn];
 inline void addedge(int u,int v)
 {
     edge[cnt].to=v;
     edge[cnt].nxt=head[u];
     head[u]=cnt++;
 }
 inline void dfs1(int u,int fa)
 {
     hvy[u]=dep[u];
     for(int e=head[u];~e;e=edge[e].nxt)
     {
         int v=edge[e].to;
         if(v==fa) continue;
         dep[v]=dep[u]+;
         dfs1(v,u);
         if(hvy[v]>hvy[u])
         {
             hson[u]=v;
             hvy[u]=hvy[v];//子树的最深深度
         }
     }
 }
 inline void dfs2(int u,int fa)
 {
     for(int e=head[u];~e;e=edge[e].nxt)
     {
         int v=edge[e].to;
         if(v==fa) continue;
         dfs2(v,u);
     }
     if(hson[u]>)
     {
         int v=hson[u];
         swap(mp[u],mp[v]);
         mp[u][dep[u]]=;
         if(maxd[v]>)
         {
             maxd[u]=maxd[v];
             ans[u]=ans[v]+;
         }
         else maxd[u]=,ans[u]=;
     }
     else maxd[u]=,ans[u]=,mp[u][dep[u]]=;
     map<int,int>::iterator it;
     for(int i=head[u];~i;i=edge[i].nxt)
     {
         int v=edge[i].to;
         if(v==fa || v==hson[u]) continue;
         for(it=mp[v].begin();it!=mp[v].end();it++)
         {
             mp[u][(*it).fi]+=mp[v][(*it).fi];
             if(mp[u][(*it).fi]>maxd[u])
             {
                 maxd[u]=mp[u][(*it).fi];
                 ans[u]=(*it).fi-dep[u];
             }
             if(mp[u][(*it).fi]==maxd[u] && (*it).fi-dep[u]<ans[u]) ans[u]=(*it).fi-dep[u];
         }
     }
 }

 int main()
 {
     n=read();int u,v;
     cnt=;clr(head,-);
     for(int i=;i<=n;++i) mp[i].clear();
     for(int i=;i<n;++i)
     {
         u=read(),v=read();
         addedge(u,v);addedge(v,u);
     }
     dep[]=;dfs1(,);dfs2(,);
     for(int i=;i<=n;++i) printf("%d%c",ans[i],i==n? '\n':' ');
     return ;
 }