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

typedef long long LL;
const int M=2e5+5;
int n;
int c[M];
vector<int> g[M];

LL ans[M];
LL sz[M];
LL cnt[M];

void dfs(int u,int fa)
{
	int nc=c[u];
	int cntu=cnt[nc];
	for(int v:g[u])
	{
		if(v==fa) continue;
		cnt[nc]=0;
		dfs(v,u);
		LL tmp=sz[v]-cnt[nc];
		ans[nc]-=tmp*(tmp-1)/2+tmp;
		sz[u]+=sz[v];
	}
	sz[u]++;
	cnt[nc]=cntu+sz[u];
}

int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
		scanf("%d",&c[i]);
	int u,v;
	for(int i=1;i<n;i++)
	{
		scanf("%d%d",&u,&v);
		g[u].push_back(v);
		g[v].push_back(u);
	}
	for(int i=1;i<=n;i++)
		ans[i]=1LL*n*(n-1)/2+n;
	dfs(1,0);
	for(int i=1;i<=n;i++)
	{
		LL tmp=sz[1]-cnt[i];
		//cout<<cnt[i]<<endl;
		ans[i]-=tmp*(tmp-1)/2+tmp;
	}
	for(int i=1;i<=n;i++)
		printf("%lld\n",ans[i]);
	return 0;
}
### 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 M=2e5+5;
int n;
int c[M];
vector<int> g[M];

LL ans[M];
LL sz[M];
LL cnt[M];

void dfs(int u,int fa)
{
	int nc=c[u];
	int cntu=cnt[nc];
	for(int v:g[u])
	{
		if(v==fa) continue;
		cnt[nc]=0;
		dfs(v,u);
		LL tmp=sz[v]-cnt[nc];
		ans[nc]-=tmp*(tmp-1)/2+tmp;
		sz[u]+=sz[v];
	}
	sz[u]++;
	cnt[nc]=cntu+sz[u];
}

int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)
		scanf("%d",&c[i]);
	int u,v;
	for(int i=1;i<n;i++)
	{
		scanf("%d%d",&u,&v);
		g[u].push_back(v);
		g[v].push_back(u);
	}
	for(int i=1;i<=n;i++)
		ans[i]=1LL*n*(n-1)/2+n;
	dfs(1,0);
	for(int i=1;i<=n;i++)
	{
		LL tmp=sz[1]-cnt[i];
		//cout<<cnt[i]<<endl;
		ans[i]-=tmp*(tmp-1)/2+tmp;
	}
	for(int i=1;i<=n;i++)
		printf("%lld\n",ans[i]);
	return 0;
}
```