E. Lomsat gelral+树上启发合并

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 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

Copy

4
1 2 3 4
1 2
2 3
2 4

Output

Copy

10 9 3 4

Input

Copy

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

Copy

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

#include<bits/stdc++.h>
using namespace std;
#define ll long long
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define all(a) (a).begin(),(a).end()
#define pll pair<ll,ll>
#define vi vector<int>
#define pb push_back
const int inf=0x3f3f3f3f;
ll rd(){
    ll 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*10+ch-'0';ch=getchar();}
    return x*f;
}
typedef map<ll,ll> mp;
const int N=1e5+10;

ll arr[N],ans[N];
mp sum[N],has[N];
vector<int> g[N];

inline void ins(mp& suml,mp& sumr,mp& hasl,mp& hasr){
    if(hasl.size()<hasr.size())swap(suml,sumr),swap(hasl,hasr);
    for(auto it:hasr){
       suml[hasl[it.first]]-=it.first;
       suml[hasl[it.first]+=it.second]+=it.first;
    }
}


void dfs(int u,int f){
    has[u][arr[u]]=1;
    sum[u][1]=arr[u];
    for(auto v:g[u]){
        if(v==f)continue;
        dfs(v,u);
        ins(sum[u],sum[v],has[u],has[v]);
    }
    ans[u]=sum[u].rbegin()->second;
}

int main(){
#ifdef happy
    freopen("in.txt","r",stdin);
#endif
    int n=rd();
    rep(i,1,n)arr[i]=rd();
    rep(i,1,n-1){
        int u=rd(),v=rd();
        g[u].push_back(v);
        g[v].push_back(u);
    }
    dfs(1,1);
    rep(i,1,n)printf("%lld ",ans[i]);
    puts("");
}