#include <iostream>
#include <vector>

using namespace std;

long long n;
vector <int> c;
vector < vector <int> > mv;
vector <long long> res;

int sol(int ac, int fa,vector <int> &el){
	long long gu = el[c[ac]];
	long long so = 1;
	long long re;
	for(int i=0;i<mv[ac].size();i++)if(mv[ac][i]!=fa){
		el[c[ac]]=0;
		re = sol(mv[ac][i],ac,el);
		so += re;
		re-=el[c[ac]];
		re = re * (re-1)/2 + re;
		res[c[ac]] -= re;
	}
	el[c[ac]] = gu +so;
	return so;
}
int main(){
	
	cin>>n;
	c.resize(n);
	for(int i=0;i<n;i++)cin>>c[i],c[i]--;
	mv.resize(n);
	res.resize(n,n*(n-1)/2+n);
	vector <int> el(n,0);
	for(int i=1;i<n;i++){
		int u,v;
		cin>>u>>v;
		u--;v--;
		mv[v].push_back(u);
		mv[u].push_back(v);
	}
	sol(0,0,el);
	for(int i=0;i<n;i++){
		long long re = n - el[i];
		re = re * (re-1)/2 + re;
		res[i] -= re;
		cout<<res[i]<<endl;
	}
	
	return 0;
}
### 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 <iostream>
#include <vector>

using namespace std;

long long n;
vector <int> c;
vector < vector <int> > mv;
vector <long long> res;

int sol(int ac, int fa,vector <int> &el){
	long long gu = el[c[ac]];
	long long so = 1;
	long long re;
	for(int i=0;i<mv[ac].size();i++)if(mv[ac][i]!=fa){
		el[c[ac]]=0;
		re = sol(mv[ac][i],ac,el);
		so += re;
		re-=el[c[ac]];
		re = re * (re-1)/2 + re;
		res[c[ac]] -= re;
	}
	el[c[ac]] = gu +so;
	return so;
}
int main(){
	
	cin>>n;
	c.resize(n);
	for(int i=0;i<n;i++)cin>>c[i],c[i]--;
	mv.resize(n);
	res.resize(n,n*(n-1)/2+n);
	vector <int> el(n,0);
	for(int i=1;i<n;i++){
		int u,v;
		cin>>u>>v;
		u--;v--;
		mv[v].push_back(u);
		mv[u].push_back(v);
	}
	sol(0,0,el);
	for(int i=0;i<n;i++){
		long long re = n - el[i];
		re = re * (re-1)/2 + re;
		res[i] -= re;
		cout<<res[i]<<endl;
	}
	
	return 0;
}
```