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+10;
int n,cnt;
int col[maxn],head[maxn],dy[maxn];
ll ans[maxn];
struct Edge{
    int to,nxt;
} edge[maxn<<1];
map<int,int> mp[maxn];
map<int,ll> sum[maxn];
map<int,int>::iterator it1,it2;
inline int read()
{
    int x=0,f=1;char ch=getchar();
    while(ch<'0' || ch>'9'){if(ch=='-') f=-1;ch=getchar();} 
    while(ch>='0' && ch<='9'){x=(x<<3)+(x<<1)+ch-'0';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=1;i<=n;++i) col[i]=read(),dy[i]=i,mp[i][col[i]]=1,sum[i][1]=col[i];
    int u,v;cnt=0;clr(head,-1);
    for(int i=1;i<n;++i) 
    {
        u=read();v=read();
        addedge(u,v);addedge(v,u);
    }
    dfs(1,0);
    for(int i=1;i<=n;++i) printf("%lld%c",ans[i],i==n? '
':' ');
    
    return 0;
}