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.
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.
Print n integers — the sums of dominating colours for each vertex.
4 1 2 3 4 1 2 2 3 2 4
10 9 3 4
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
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(""); }