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 ci (1 ≤ ci ≤ n), ci — the colour of the i-th vertex.

Each of the next n - 1 lines contains two integers xj, yj (1 ≤ xj, yj ≤ 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
import java.io.BufferedReader;
import java.io.BufferedWriter;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStream;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.io.StreamTokenizer;
import java.math.BigDecimal;
import java.math.BigInteger;
import java.util.Arrays;
import java.util.HashMap;
import java.util.HashSet;
import java.util.LinkedList;
import java.util.Map;
import java.util.Scanner;
import java.util.Set;
import java.util.Stack;
import java.util.TreeSet;



public class Main
{
	static int sz[],size[],son[],max,cnt[],o,p;
	static long ans[],sum[];
	static int top[],next[],end[];
	static void con(int a,int b)
	{
		next[++o]=top[a];
		top[a]=o;
		end[o]=b;
	}
	static void dfs(int x,int fu)
	{
		size[x]=1;
		for (int i=top[x];i!=0;i=next[i])
		{
			if (end[i]!=fu)
			{
				dfs(end[i],x);
				size[x]+=size[end[i]];
				if (size[son[x]]<size[end[i]])son[x]=end[i];
			}
		}
	}
	static void g(int x,int fu,int s)
	{
		sum[cnt[sz[x]]]-=sz[x];//cnt[sz[x]]默认是0的，我们不使用索引号0的位置，所以0位置可以为负数
		cnt[sz[x]]+=s;
		sum[cnt[sz[x]]]+=sz[x];
		if (sum[max+1]>0)max++;
		if (sum[max]==0)max--;
		for (int i=top[x];i!=0;i=next[i])
		{
			if (end[i]!=p && end[i]!=fu)//因为重节点的值是保留的，所以end[i]!=p避免重结点的值重复操作
			{
				g(end[i],x,s);
			}
		}
		
	}
	static void dfs1(int x,int fu,int s)//当前结点，父结点，s==1代表重结点
	{
		
		for (int i=top[x];i!=0;i=next[i])
		{
			if (end[i]!=son[x] && end[i]!=fu)
			{
				dfs1(end[i],x,0);
			}
		}
		if (son[x]!=0)//如果不是叶子结点，就有重儿子
		{
			dfs1(son[x],x,1);
			p=son[x];//标记重结点，免得dfs求值重结点重复记录
		}
		g(x,fu,1);//dfs一遍该结点的子结点
		p=0;
		ans[x]=sum[max];//记录答案
		if (s==0)g(x,fu,-1);//如果不是重结点，删除数组
	}
	public static void main(String[] args) throws IOException
	{
		int n=ini();
		p=0;o=0;
		sz=new int [n+1];
		ans=new long [n+1];
		
		top=new int [n+1];
		next=new int [n<<1];
		end=new int [n<<1];
		
		size=new int [n+1];
		son=new int [n+1];
		sum=new long [n+1];
		cnt=new int [n+1];
		for (int i=1;i<=n;i++)sz[i]=ini();
		for (int i=1;i<n;i++)
		{
			int a=ini();
			int b=ini();
			con(a,b);
			con(b,a);
		}
		dfs(1,0);
		dfs1(1,0,1);
		for (int i=1;i<=n;i++)
		{
			if (i==1)out.print(ans[i]);
			else out.print(" "+ans[i]);
		}
		
		out.flush();
	}
/*


*/
	static StreamTokenizer in=new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
	static PrintWriter out=new PrintWriter(new BufferedWriter(new OutputStreamWriter(System.out)));
	static int ini() throws IOException
	{
		in.nextToken();
		return (int)in.nval;
	}
	static long inl() throws IOException
	{
		in.nextToken();
		return (long)in.nval;
	}
	static String ins() throws IOException
	{
		in.nextToken();
		return in.sval;
	}
	
}