#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
const int N = 2e5;
int c[N+100],subt[N+100],ind[N+100],par[N+100];
vector<int>g[N+100],com[N+100],col[N+100];
vector<int>lst[N+100];
ll ans[N+100];

void dfs(int v,int Par)
{
	int cur = c[v];
	if(!lst[cur].empty())
	{
		int u = lst[cur].back();
		com[ind[u]].push_back(v);
	}
	else{
		col[cur].push_back(v);
	}
	lst[cur].push_back(v);
	subt[v] = 1;
	for(auto it:g[v])
	{
		if(it!=Par){
			ind[v] = it;
			par[it] = v;
			dfs(it,v);
			subt[v] += subt[it];
		}
	}
	lst[cur].pop_back();
}
int main()
{
	int n;
	scanf("%d",&n);
	for(int i=1;i<=n;i++){
		scanf("%d",c+i);
		ans[i] = (1ll*n*(n+1))/2;
	}
	for(int i=1;i<n;i++){
		int a,b;
		scanf("%d%d",&a,&b);
		g[a].push_back(b);
		g[b].push_back(a);
	}
	dfs(1,0);
	for(int i=1;i<=n;i++){
		int sum = n;
		for(auto it:col[i]){
			sum -= subt[it];
		}
		ans[i] -= (1ll*sum*(sum+1))/2;
		//printf("%lld\n",ans[i]);
	}
	for(int i=2;i<=n;i++){
		int pre = c[par[i]];
		int cur = subt[i];
		for(auto it:com[i]){
			cur -= subt[it];
		}
		ans[pre] -= (1ll*cur*(cur+1))/2;
	}
	for(int i=1;i<=n;i++){
		printf("%lld\n",ans[i]);
	}
}

### Prompt

Please provide a cpp coded solution to the problem described below:

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;

typedef long long ll;
const int N = 2e5;
int c[N+100],subt[N+100],ind[N+100],par[N+100];
vector<int>g[N+100],com[N+100],col[N+100];
vector<int>lst[N+100];
ll ans[N+100];

void dfs(int v,int Par)
{
	int cur = c[v];
	if(!lst[cur].empty())
	{
		int u = lst[cur].back();
		com[ind[u]].push_back(v);
	}
	else{
		col[cur].push_back(v);
	}
	lst[cur].push_back(v);
	subt[v] = 1;
	for(auto it:g[v])
	{
		if(it!=Par){
			ind[v] = it;
			par[it] = v;
			dfs(it,v);
			subt[v] += subt[it];
		}
	}
	lst[cur].pop_back();
}
int main()
{
	int n;
	scanf("%d",&n);
	for(int i=1;i<=n;i++){
		scanf("%d",c+i);
		ans[i] = (1ll*n*(n+1))/2;
	}
	for(int i=1;i<n;i++){
		int a,b;
		scanf("%d%d",&a,&b);
		g[a].push_back(b);
		g[b].push_back(a);
	}
	dfs(1,0);
	for(int i=1;i<=n;i++){
		int sum = n;
		for(auto it:col[i]){
			sum -= subt[it];
		}
		ans[i] -= (1ll*sum*(sum+1))/2;
		//printf("%lld\n",ans[i]);
	}
	for(int i=2;i<=n;i++){
		int pre = c[par[i]];
		int cur = subt[i];
		for(auto it:com[i]){
			cur -= subt[it];
		}
		ans[pre] -= (1ll*cur*(cur+1))/2;
	}
	for(int i=1;i<=n;i++){
		printf("%lld\n",ans[i]);
	}
}

```