#include<bits/stdc++.h>
#define ll long long
#define vi vector<int>
#define pb push_back
using namespace std;
const int N=2e5+5;
int sum[N],col[N],n;
ll ans[N],son[N];
vi v[N];
void DFS(int now,int fa)
{
	son[now]=1;
	for(int i=0;i<v[now].size();i++)
	{
		int to=v[now][i];
		if(to==fa)continue;
		DFS(to,now);
		son[now]+=son[to];
	}
}
void dfs(int now,int fa)
{
	ans[col[now]]+=(ll)son[now]*(ll)(n-son[now]-(ll)sum[col[now]]+1);
	sum[col[now]]+=son[now];
	int rem=sum[col[now]];
	ll pre=0;
	for(int i=0;i<v[now].size();i++)
	{
		int to=v[now][i];
		if(to==fa)continue;
		sum[col[now]]=(n-son[to]);
		dfs(to,now);
		ans[col[now]]+=pre*son[to];
		pre+=son[to];
	}
	sum[col[now]]=rem;
	return;
}
int main()
{
	cin>>n;
	for(int i=0;i<n;i++)cin>>col[i];
	for(int i=0;i<n-1;i++)
	{
		int a,b;
		cin>>a>>b;
		a--,b--;
		v[a].pb(b);
		v[b].pb(a);
	}
	DFS(0,-1);
	dfs(0,-1);
	for(int i=1;i<=n;i++)cout<<ans[i]<<"\n";
	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<bits/stdc++.h>
#define ll long long
#define vi vector<int>
#define pb push_back
using namespace std;
const int N=2e5+5;
int sum[N],col[N],n;
ll ans[N],son[N];
vi v[N];
void DFS(int now,int fa)
{
	son[now]=1;
	for(int i=0;i<v[now].size();i++)
	{
		int to=v[now][i];
		if(to==fa)continue;
		DFS(to,now);
		son[now]+=son[to];
	}
}
void dfs(int now,int fa)
{
	ans[col[now]]+=(ll)son[now]*(ll)(n-son[now]-(ll)sum[col[now]]+1);
	sum[col[now]]+=son[now];
	int rem=sum[col[now]];
	ll pre=0;
	for(int i=0;i<v[now].size();i++)
	{
		int to=v[now][i];
		if(to==fa)continue;
		sum[col[now]]=(n-son[to]);
		dfs(to,now);
		ans[col[now]]+=pre*son[to];
		pre+=son[to];
	}
	sum[col[now]]=rem;
	return;
}
int main()
{
	cin>>n;
	for(int i=0;i<n;i++)cin>>col[i];
	for(int i=0;i<n-1;i++)
	{
		int a,b;
		cin>>a>>b;
		a--,b--;
		v[a].pb(b);
		v[b].pb(a);
	}
	DFS(0,-1);
	dfs(0,-1);
	for(int i=1;i<=n;i++)cout<<ans[i]<<"\n";
	return 0;
}
```