Problem:
Score : 600 points

Problem Statement

We have a tree with N vertices numbered 1 to N. The i-th edge in this tree connects Vertex a_i and b_i.
Additionally, each vertex is painted in a color, and the color of Vertex i is c_i. Here, the color of each vertex is represented by an integer between 1 and N (inclusive). The same integer corresponds to the same color; different integers correspond to different colors.

For each k=1, 2, ..., N, solve the following problem:

Find the number of simple paths that visit a vertex painted in the color k one or more times.

Note: The simple paths from Vertex u to v and from v to u are not distinguished.

Constraints

1 \leq N \leq 2 \times 10^5

1 \leq c_i \leq N

1 \leq a_i,b_i \leq N

The given graph is a tree.

All values in input are integers.

Input

Input is given from Standard Input in the following format:

N
c_1 c_2 ... c_N
a_1 b_1
:
a_{N-1} b_{N-1}

Output

Print the answers for k = 1, 2, ..., N in order, each in its own line.

Sample Input 1

3
1 2 1
1 2
2 3

Sample Output 1

5
4
0

Let P_{i,j} denote the simple path connecting Vertex i and j.

There are 5 simple paths that visit a vertex painted in the color 1 one or more times:

P_{1,1}\,,\,
P_{1,2}\,,\,
P_{1,3}\,,\,
P_{2,3}\,,\,
P_{3,3}

There are 4 simple paths that visit a vertex painted in the color 2 one or more times:

P_{1,2}\,,\,
P_{1,3}\,,\,
P_{2,2}\,,\,
P_{2,3}

There are no simple paths that visit a vertex painted in the color 3 one or more times.

Sample Input 2

1
1

Sample Output 2

1

Sample Input 3

2
1 2
1 2

Sample Output 3

2
2

Sample Input 4

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

Sample Output 4

5
8
10
5
5

Sample Input 5

8
2 7 2 5 4 1 7 5
3 1
1 2
2 7
4 5
5 6
6 8
7 8

Sample Output 5

18
15
0
14
23
0
23
0

Solution in C++ (GCC 5.4.1) (judge verdict: Accepted)

/*
读题不规范，爆零两行泪。
数据不清空，爆零两行泪。
多测不读完，爆零两行泪。
边界不特判，爆零两行泪。
贪心不证明，爆零两行泪。
D P 顺序错，爆零两行泪。
大小少等号，爆零两行泪。
变量不统一，爆零两行泪。
越界不判断，爆零两行泪。
调试不注释，爆零两行泪。
溢出不 l l，爆零两行泪。
*/
#include<bits/stdc++.h>
using namespace std;
#define pb push_back
#define ppb pop_back
typedef long long ll;
const int N=200000;
int n;//节点个数 
int a[N+1];//颜色 
vector<int> nei[N+1];//邻接表 
int sz[N+1]/*子树大小*/,cc_sz[N+1]/*CC大小（非根）*/,cc_sz1[N+1]/*CC大小（根，下标为颜色）*/;
int fa[N+1];//父亲 
vector<int> stk/*总递归栈*/,stk_c[N+1]/*颜色们的递归栈*/;
void dfs(int x=1){//求CC们的大小 
//	cout<<x<<"\n";
	sz[x]=1;
	stk_c[a[x]].pb(stk.size());
	stk.pb(x);//压栈 
	for(int i=0;i<nei[x].size();i++){
		int y=nei[x][i];
		if(y==fa[x])continue;
		fa[y]=x;
		dfs(y);
		sz[x]+=sz[y];
	}
	stk.ppb();
	stk_c[a[x]].ppb();//准备回溯，弹出 
	if(stk_c[a[x]].size()>1)cc_sz[stk[stk_c[a[x]].back()+1]]-=sz[x];//贡献（非根） 
	else/*特判*/ cc_sz1[a[x]]-=sz[x];//贡献（根） 
}
ll ans[N+1];
int main(){
	cin>>n;
	for(int i=1;i<=n;i++)cin>>a[i];
	for(int i=1;i<n;i++){
		int x,y;
		cin>>x>>y;
		nei[x].pb(y);nei[y].pb(x);
	}
	stk.pb(0);
	for(int i=1;i<=n;i++)stk_c[i].pb(0);
	dfs();
	for(int i=1;i<=n;i++)cc_sz[i]+=sz[i],cc_sz1[i]+=n;
//	for(int i=1;i<=n;i++)cout<<cc_sz[i]<<" ";puts("");
//	for(int i=1;i<=n;i++)cout<<cc_sz1[i]<<" ";puts("");
	for(int i=1;i<=n;i++)ans[i]=1ll*n*(n+1)/2;
	for(int i=2;i<=n;i++)ans[a[fa[i]]]-=1ll*cc_sz[i]*(cc_sz[i]+1)/2;
	for(int i=1;i<=n;i++)ans[i]-=1ll*cc_sz1[i]*(cc_sz1[i]+1)/2;//减去反面答案 
	for(int i=1;i<=n;i++)cout<<ans[i]<<"\n";
	return 0;
}