#include<iostream>
#include<vector>
using namespace std;
int N;
int c[2<<17];
vector<int>G[2<<17];
int cnt[2<<17];
long ans[2<<17];
int dfs(int u,int p)
{
	int ret=1;
	int sto=cnt[c[u]];
	int pre=sto;
	for(int v:G[u])if(v!=p)
	{
		int t=dfs(v,u);
		ret+=t;
		long k=t-(cnt[c[u]]-sto);
		ans[c[u]]-=k*(k+1)/2;
		sto=cnt[c[u]];
	}
	cnt[c[u]]=pre+ret;
	return ret;
}
int main()
{
	cin>>N;
	for(int i=0;i<N;i++)
	{
		cin>>c[i];
		c[i]--;
	}
	for(int i=1;i<N;i++)
	{
		int u,v;cin>>u>>v;u--,v--;
		G[u].push_back(v);
		G[v].push_back(u);
	}
	dfs(0,-1);
	for(int i=0;i<N;i++)
	{
		ans[i]+=(long)N*(N+1)/2;
		long k=N-cnt[i];
		ans[i]-=k*(k+1)/2;
		cout<<ans[i]<<endl;
	}
}
### Prompt

Please formulate 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;
int N;
int c[2<<17];
vector<int>G[2<<17];
int cnt[2<<17];
long ans[2<<17];
int dfs(int u,int p)
{
	int ret=1;
	int sto=cnt[c[u]];
	int pre=sto;
	for(int v:G[u])if(v!=p)
	{
		int t=dfs(v,u);
		ret+=t;
		long k=t-(cnt[c[u]]-sto);
		ans[c[u]]-=k*(k+1)/2;
		sto=cnt[c[u]];
	}
	cnt[c[u]]=pre+ret;
	return ret;
}
int main()
{
	cin>>N;
	for(int i=0;i<N;i++)
	{
		cin>>c[i];
		c[i]--;
	}
	for(int i=1;i<N;i++)
	{
		int u,v;cin>>u>>v;u--,v--;
		G[u].push_back(v);
		G[v].push_back(u);
	}
	dfs(0,-1);
	for(int i=0;i<N;i++)
	{
		ans[i]+=(long)N*(N+1)/2;
		long k=N-cnt[i];
		ans[i]-=k*(k+1)/2;
		cout<<ans[i]<<endl;
	}
}
```