#include <bits/stdc++.h>
using namespace std;
long long cur;
vector<int> inv[200005],v[200005];
int stt[200005],ent[200005],sz[200005],ct;
set<pair<int,int> > s;
int dfs(int node,int p)
{
	sz[node]=1;
	stt[node]=++ct;
	for (int u:v[node])
	{
		if (u!=p)
		sz[node]+=dfs(u,node);
	}
	ent[node]=ct;
	return sz[node];
}
void dfs2(int node);
void dfs3(int node)
{
	long long cmp=sz[node];
	while (1)
	{
		auto it=s.lower_bound({stt[node],node});
		if (it==s.end() || it->first>ent[node])
		break;
		cmp-=sz[it->second];
		dfs2(it->second);
	}
	cur-=cmp*(cmp+1)/2;
}
void dfs2(int node)
{
	s.erase({stt[node],node});
	for (int u:v[node])
	{
		if (stt[u]>=stt[node])
		dfs3(u);
	}
}
int main()
{
	int n;
	scanf("%d",&n);
	for (int i=1;i<=n;i++)
	{
		int c;
		scanf("%d",&c);
		inv[c].push_back(i);
	}
	for (int i=1;i<n;i++)
	{
		int a,b;
		scanf("%d%d",&a,&b);
		v[a].push_back(b);
		v[b].push_back(a);
	}
	v[0].push_back(1);
	dfs(0,0);
	for (int i=1;i<=n;i++)
	{
		s.clear();
		s.insert({1,0});
		for (int u:inv[i])
		s.insert({stt[u],u});
		cur=1LL*n*(n+1)/2;
		dfs2(0);
		printf("%lld\n",cur);
	}
}
### 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;
long long cur;
vector<int> inv[200005],v[200005];
int stt[200005],ent[200005],sz[200005],ct;
set<pair<int,int> > s;
int dfs(int node,int p)
{
	sz[node]=1;
	stt[node]=++ct;
	for (int u:v[node])
	{
		if (u!=p)
		sz[node]+=dfs(u,node);
	}
	ent[node]=ct;
	return sz[node];
}
void dfs2(int node);
void dfs3(int node)
{
	long long cmp=sz[node];
	while (1)
	{
		auto it=s.lower_bound({stt[node],node});
		if (it==s.end() || it->first>ent[node])
		break;
		cmp-=sz[it->second];
		dfs2(it->second);
	}
	cur-=cmp*(cmp+1)/2;
}
void dfs2(int node)
{
	s.erase({stt[node],node});
	for (int u:v[node])
	{
		if (stt[u]>=stt[node])
		dfs3(u);
	}
}
int main()
{
	int n;
	scanf("%d",&n);
	for (int i=1;i<=n;i++)
	{
		int c;
		scanf("%d",&c);
		inv[c].push_back(i);
	}
	for (int i=1;i<n;i++)
	{
		int a,b;
		scanf("%d%d",&a,&b);
		v[a].push_back(b);
		v[b].push_back(a);
	}
	v[0].push_back(1);
	dfs(0,0);
	for (int i=1;i<=n;i++)
	{
		s.clear();
		s.insert({1,0});
		for (int u:inv[i])
		s.insert({stt[u],u});
		cur=1LL*n*(n+1)/2;
		dfs2(0);
		printf("%lld\n",cur);
	}
}
```