Educational Codeforces Round 2 E. Lomsat gelral 启发式合并map

E. Lomsat gelral

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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

题目链接：http://codeforces.com/contest/600/problem/E
题意：给你一颗树，找出每个子树下的众数之和；

思路：启发式合并map；

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll long long
#define pi (4*atan(1.0))
#define eps 1e-14
#define bug(x)  cout<<"bug"<<x<<endl;
const int N=1e5+10,M=4e6+10,inf=2147483647;
const ll INF=1e18+10,mod=1e9+7;
struct is
{
    int v,next;
}edge[N<<1];
int head[N],edg;
void init()
{
    edg=0;
    memset(head,-1,sizeof(head));
}
void add(int u,int v)
{
    edg++;
    edge[edg].v=v;
    edge[edg].next=head[u];
    head[u]=edg;
}
ll ans[N],sum[N];
int a[N];
int maxx[N];
map<int,int>mp[N];
map<int,int>::iterator it;

void update(int a,int b)
{
    if(mp[a].size()<mp[b].size())
    {
        swap(maxx[a],maxx[b]);
        swap(sum[a],sum[b]);
        swap(mp[a],mp[b]);
    }
    while(mp[b].size())
    {
        it=mp[b].begin();
        mp[a][it->first]+=it->second;
        if(maxx[a]==mp[a][it->first])
            sum[a]+=it->first;
        else if(maxx[a]<mp[a][it->first])
        {
            maxx[a]=mp[a][it->first];
            sum[a]=it->first;
        }
        mp[b].erase(it);
    }
}

void dfs(int u,int fa)
{
    //cout<<u<<" "<<fa<<endl;
    mp[u][a[u]]++;
    maxx[u]=1;
    sum[u]=a[u];
    for(int i=head[u];i!=-1;i=edge[i].next)
    {
        int v=edge[i].v;
        if(v==fa)continue;
        dfs(v,u);
        update(u,v);
    }
    ans[u]=sum[u];
}
///   数组大小
int main()
{
    init();
    int n;
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]);
    for(int i=1;i<n;i++)
    {
        int u,v;
        scanf("%d%d",&u,&v);
        add(u,v),add(v,u);
    }
    dfs(1,-1);
    for(int i=1;i<=n;i++)
        printf("%lld ",ans[i]);
    return 0;
}

E. Lomsat gelral

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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