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

const int N=400005;
int n,a[N],sum[N];
int v[N],nxt[N],head[N];
long long sz[N],ans[N];

void dfs(int x,int fa)
{
	sz[x]=1;
	int ts=sum[a[x]];
	for(int i=head[x];i>0;i=nxt[i])
	{
		if(v[i]==fa) continue;
		int tmp=sum[a[x]];
		dfs(v[i],x);
		int d=(sz[v[i]]-sum[a[x]]+tmp);
		ans[a[x]]+=1ll*d*(d+1)/2;
		sz[x]+=sz[v[i]];
	}
	sum[a[x]]=ts+sz[x];
}

int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++) scanf("%d",&a[i]);
	for(int i=1;i<n;i++)
	{
		scanf("%d%d",&v[i],&v[i+n]);
		nxt[i]=head[v[i+n]],head[v[i+n]]=i;
		nxt[i+n]=head[v[i]],head[v[i]]=i+n;
	}
	dfs(1,0);
//	for(int i=1;i<=n;i++) cout<<ans[i]<<' ';
//	cout<<endl;
//	cout<<sz[3]<<endl;
	for(int i=1;i<=n;i++)
	{
//		cout<<ans[i]<<' '<<sum[i]<<endl;
		ans[i]+=1ll*(n-sum[i])*(n-sum[i]+1)/2;
		printf("%lld\n",1ll*n*(n+1)/2-ans[i]);
	}
}
### Prompt

Create a solution in CPP for 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>
using namespace std;

const int N=400005;
int n,a[N],sum[N];
int v[N],nxt[N],head[N];
long long sz[N],ans[N];

void dfs(int x,int fa)
{
	sz[x]=1;
	int ts=sum[a[x]];
	for(int i=head[x];i>0;i=nxt[i])
	{
		if(v[i]==fa) continue;
		int tmp=sum[a[x]];
		dfs(v[i],x);
		int d=(sz[v[i]]-sum[a[x]]+tmp);
		ans[a[x]]+=1ll*d*(d+1)/2;
		sz[x]+=sz[v[i]];
	}
	sum[a[x]]=ts+sz[x];
}

int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++) scanf("%d",&a[i]);
	for(int i=1;i<n;i++)
	{
		scanf("%d%d",&v[i],&v[i+n]);
		nxt[i]=head[v[i+n]],head[v[i+n]]=i;
		nxt[i+n]=head[v[i]],head[v[i]]=i+n;
	}
	dfs(1,0);
//	for(int i=1;i<=n;i++) cout<<ans[i]<<' ';
//	cout<<endl;
//	cout<<sz[3]<<endl;
	for(int i=1;i<=n;i++)
	{
//		cout<<ans[i]<<' '<<sum[i]<<endl;
		ans[i]+=1ll*(n-sum[i])*(n-sum[i]+1)/2;
		printf("%lld\n",1ll*n*(n+1)/2-ans[i]);
	}
}
```