CF600E Lomsat gelral (启发式合并)

You are given a rooted tree with root in vertex 1. Each vertex is coloured in some colour.

Let's call colour c dominating in the subtree of vertex v if there are no other colours that appear in the subtree of vertex v more times than colour c. So it's possible that two or more colours will be dominating in the subtree of some vertex.

The subtree of vertex v is the vertex v and all other vertices that contains vertex v in each path to the root.

For each vertex v find the sum of all dominating colours in the subtree of vertex v.

Input

The first line contains integer n (1 ≤ n ≤ 10⁵) — the number of vertices in the tree.

The second line contains n integers c_i (1 ≤ c_i ≤ n), c_i — the colour of the i-th vertex.

Each of the next n - 1 lines contains two integers x_j, y_j (1 ≤ x_j, y_j ≤ n) — the edge of the tree. The first vertex is the root of the tree.

Output

Print n integers — the sums of dominating colours for each vertex.

Examples

Input

4
1 2 3 4
1 2
2 3
2 4

Output

10 9 3 4

Input

15
1 2 3 1 2 3 3 1 1 3 2 2 1 2 3
1 2
1 3
1 4
1 14
1 15
2 5
2 6
2 7
3 8
3 9
3 10
4 11
4 12
4 13

Output

6 5 4 3 2 3 3 1 1 3 2 2 1 2 3

题解：题意就是定义一个节点的优势点为他的子树中出现次数最多的数字，（可能会有多个优势点），让你求每一个点的优势点的和；
可以用map记录每一个点的各个优势点，记录每点的子树中每个数字出现的次数，然后dfs。
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=1e5+;
 int n,cnt;
 int col[maxn],head[maxn],dy[maxn];
 ll ans[maxn];
 struct Edge{
     int to,nxt;
 } edge[maxn<<];
 map<int,int> mp[maxn];
 map<int,ll> sum[maxn];
 map<int,int>::iterator it1,it2;
 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;
 }

 inline void addedge(int u,int v)
 {
     edge[cnt].to=v;
     edge[cnt].nxt=head[u];
     head[u]=cnt++;
 }

 inline void dfs(int u,int fa)
 {
     for(int e=head[u];~e;e=edge[e].nxt)
     {
         int v=edge[e].to;
         if(v==fa) continue;
         dfs(v,u);
         if(mp[dy[u]].size()<mp[dy[v]].size()) swap(dy[u],dy[v]);
         for(it1=mp[dy[v]].begin();it1!=mp[dy[v]].end();it1++)
         {
             sum[dy[u]][mp[dy[u]][(*it1).fi]]-=(*it1).fi;
             mp[dy[u]][(*it1).fi]+=(*it1).se;
             sum[dy[u]][mp[dy[u]][(*it1).fi]]+=(*it1).fi;
         }
     }
     ans[u]=(*sum[dy[u]].rbegin()).se;
 }

 int main()
 {
     n=read();
     for(int i=;i<=n;++i) col[i]=read(),dy[i]=i,mp[i][col[i]]=,sum[i][]=col[i];
     int u,v;cnt=;clr(head,-);
     for(int i=;i<n;++i)
     {
         u=read();v=read();
         addedge(u,v);addedge(v,u);
     }
     dfs(,);
     for(int i=;i<=n;++i) printf("%lld%c",ans[i],i==n? '\n':' ');

     return ;
 }