Dominant Indices

题目

You are given a rooted undirected tree consisting of n vertices. Vertex 1 is the root.

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

x is an ancestor of y;
the simple path from x to y traverses exactly i edges.
The dominant index of a depth array of vertex x (or, shortly, the dominant index of vertex x) is an index j such that:

for every k<j, dx,k<dx,j;
for every k>j, dx,k≤dx,j.
For every vertex in the tree calculate its dominant index.

Input

The first line contains one integer n (1≤n≤10^6) — the number of vertices in a tree.

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

It is guaranteed that these edges form a tree.

Output

Output n numbers. i-th number should be equal to the dominant index of vertex i.

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

思路

和上面一道题目差不多,都是树上启发式合并,就是统计一下深度就好了啦~

代码如下

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<climits>
#include<cctype>
#include<queue>
#include<set>
using namespace std;
typedef long long ll;
const int INF=0x3f3f3f3f;
const int mx=1e6+10;
int n,head[mx],tot;
int size[mx],deep[mx],son[mx];
int cnt[mx],nowval,maxval,vis[mx],ans[mx];struct node{int to,next;
}e[mx<<1];void init()
{tot=0;memset(head,-1,sizeof(head));memset(vis,0,sizeof(vis));
}void add(int u,int v)
{e[tot].to=v;e[tot].next=head[u];head[u]=tot++;
}void dfs(int u,int fa,int d)
{size[u]=1;deep[u]=d;for(int i=head[u];i!=-1;i=e[i].next){int v=e[i].to;if(v==fa) continue;dfs(v,u,d+1);size[u]+=size[v];if(size[v]>size[son[u]])son[u]=v;}
}void update(int u,int fa,int k)
{cnt[deep[u]]+=k;if(k>0&&cnt[deep[u]]>=maxval){if(cnt[deep[u]]>maxval) maxval=cnt[deep[u]],nowval=deep[u];else if(cnt[deep[u]]==maxval&&deep[u]<nowval)nowval=deep[u];}for(int i=head[u];i!=-1;i=e[i].next){int v=e[i].to;if(v!=fa&&!vis[v])update(v,u,k);}
}void dsu(int u,int fa,int keep)
{for(int i=head[u];i!=-1;i=e[i].next){int v=e[i].to;if(v!=fa&&v!=son[u])//优先轻儿子dsu(v,u,0);}if(son[u]) dsu(son[u],u,1),vis[son[u]]=1;update(u,fa,1);ans[u]=nowval-deep[u];if(son[u]) vis[son[u]]=0;if(!keep) update(u,fa,-1),nowval=maxval=0;
}int main()
{init();scanf("%d",&n);for(int i=1;i<n;i++){int u,v;scanf("%d%d",&u,&v);add(u,v);add(v,u);}dfs(1,0,1);dsu(1,0,1);for(int i=1;i<=n;i++)printf("%d\n",ans[i]);return 0;
}