#include<bits/stdc++.h>
using namespace std;
using ll=long long;
constexpr ll mod=998244353;

ll res[200000];
int N,c[200000],s[200000],p[200000],t[200000];
vector<int>v[200000],k[200000];
stack<int>st[200000];

void dfs(int x,int from){
	p[x]=from;
	s[x]=1;
	if(from!=-1){
		st[c[from]].push(x);
	}
	if(!st[c[x]].empty())k[st[c[x]].top()].push_back(x);

	for(int i:v[x]){
		if(i==from)continue;
		dfs(i,x);
		s[x]+=s[i];
	}
	if(st[c[x]].empty()){
		t[c[x]]-=s[x];
	}
	if(from!=-1){
		st[c[from]].pop();
	}

}

int main(){
	cin>>N;
	for(int i=0;i<N;i++){
		cin>>c[i];
		c[i]--;
	}
	for(int i=1;i<N;i++){
		int a,b;
		cin>>a>>b;
		a--;
		b--;
		v[a].push_back(b);
		v[b].push_back(a);
	}
	for(int i=0;i<N;i++){
		t[i]=N;
		res[i]=(ll)N*(N+1)/2;
	}
	dfs(0,-1);
	for(int i=0;i<N;i++){
		res[i]-=(ll)t[i]*(t[i]+1)/2;
	//	cout<<t[i]<<endl;
	}
	for(int i=1;i<N;i++){
		int a=s[i];
		for(int j:k[i]){
			a-=s[j];
		}
		res[c[p[i]]]-=(ll)a*(a+1)/2;
	}
	for(int i=0;i<N;i++){
		cout<<res[i]<<endl;
	}
}

### Prompt

Your task is to create a cpp solution to the following problem:

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.

Examples

Input

3
1 2 1
1 2
2 3


Output

5
4
0


Input

1
1


Output

1


Input

2
1 2
1 2


Output

2
2


Input

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


Output

5
8
10
5
5


Input

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


Output

18
15
0
14
23
0
23
0

### Response

```cpp
#include<bits/stdc++.h>
using namespace std;
using ll=long long;
constexpr ll mod=998244353;

ll res[200000];
int N,c[200000],s[200000],p[200000],t[200000];
vector<int>v[200000],k[200000];
stack<int>st[200000];

void dfs(int x,int from){
	p[x]=from;
	s[x]=1;
	if(from!=-1){
		st[c[from]].push(x);
	}
	if(!st[c[x]].empty())k[st[c[x]].top()].push_back(x);

	for(int i:v[x]){
		if(i==from)continue;
		dfs(i,x);
		s[x]+=s[i];
	}
	if(st[c[x]].empty()){
		t[c[x]]-=s[x];
	}
	if(from!=-1){
		st[c[from]].pop();
	}

}

int main(){
	cin>>N;
	for(int i=0;i<N;i++){
		cin>>c[i];
		c[i]--;
	}
	for(int i=1;i<N;i++){
		int a,b;
		cin>>a>>b;
		a--;
		b--;
		v[a].push_back(b);
		v[b].push_back(a);
	}
	for(int i=0;i<N;i++){
		t[i]=N;
		res[i]=(ll)N*(N+1)/2;
	}
	dfs(0,-1);
	for(int i=0;i<N;i++){
		res[i]-=(ll)t[i]*(t[i]+1)/2;
	//	cout<<t[i]<<endl;
	}
	for(int i=1;i<N;i++){
		int a=s[i];
		for(int j:k[i]){
			a-=s[j];
		}
		res[c[p[i]]]-=(ll)a*(a+1)/2;
	}
	for(int i=0;i<N;i++){
		cout<<res[i]<<endl;
	}
}

```