#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
using namespace std;
#define ll long long
struct data_p{int hed,clr;ll siz;}pot[200005];
struct data_l{int to,nxt;}lin[400005];
struct data_q{ll now,ans;}qst[200005];
ll n;
int top=0;
void add_l(int a,int b){lin[++top].to=b;lin[top].nxt=pot[a].hed;pot[a].hed=top;}
void get_siz(int a,int fa)
{
	pot[a].siz=1;int b;
	for(int i=pot[a].hed;i;i=lin[i].nxt)
	{
		b=lin[i].to;
		if(b==fa)continue;
		get_siz(b,a);
		pot[a].siz+=pot[b].siz;
	}
}
void dp(int a,int fa)
{
	ll tmp=qst[pot[a].clr].now,potl=n-pot[a].siz+1-qst[pot[a].clr].now;int b;
	qst[pot[a].clr].ans+=potl;
	for(int i=pot[a].hed;i;i=lin[i].nxt)
	{
		b=lin[i].to;
		if(b==fa)continue;
		qst[pot[a].clr].now=n-pot[b].siz;
		dp(b,a);
		qst[pot[a].clr].ans+=potl*pot[b].siz;
		potl+=pot[b].siz;
	}
	qst[pot[a].clr].now=tmp+pot[a].siz;
}
int main()
{
	scanf("%lld",&n);
	for(int i=1;i<=n;i++)
	scanf("%d",&pot[i].clr);
	for(int i=1;i<n;i++)
	{
		int a,b;scanf("%d%d",&a,&b);
		add_l(a,b);add_l(b,a);
	}
	get_siz(1,1);
	dp(1,1);
	for(int i=1;i<=n;i++)
	{
		printf("%lld\n",qst[i].ans);
	}
	return 0;
}
### Prompt

Your challenge is to write 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<cstdio>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<cstring>
using namespace std;
#define ll long long
struct data_p{int hed,clr;ll siz;}pot[200005];
struct data_l{int to,nxt;}lin[400005];
struct data_q{ll now,ans;}qst[200005];
ll n;
int top=0;
void add_l(int a,int b){lin[++top].to=b;lin[top].nxt=pot[a].hed;pot[a].hed=top;}
void get_siz(int a,int fa)
{
	pot[a].siz=1;int b;
	for(int i=pot[a].hed;i;i=lin[i].nxt)
	{
		b=lin[i].to;
		if(b==fa)continue;
		get_siz(b,a);
		pot[a].siz+=pot[b].siz;
	}
}
void dp(int a,int fa)
{
	ll tmp=qst[pot[a].clr].now,potl=n-pot[a].siz+1-qst[pot[a].clr].now;int b;
	qst[pot[a].clr].ans+=potl;
	for(int i=pot[a].hed;i;i=lin[i].nxt)
	{
		b=lin[i].to;
		if(b==fa)continue;
		qst[pot[a].clr].now=n-pot[b].siz;
		dp(b,a);
		qst[pot[a].clr].ans+=potl*pot[b].siz;
		potl+=pot[b].siz;
	}
	qst[pot[a].clr].now=tmp+pot[a].siz;
}
int main()
{
	scanf("%lld",&n);
	for(int i=1;i<=n;i++)
	scanf("%d",&pot[i].clr);
	for(int i=1;i<n;i++)
	{
		int a,b;scanf("%d%d",&a,&b);
		add_l(a,b);add_l(b,a);
	}
	get_siz(1,1);
	dp(1,1);
	for(int i=1;i<=n;i++)
	{
		printf("%lld\n",qst[i].ans);
	}
	return 0;
}
```