#include<bits/stdc++.h>
using namespace std;
#define int long long 
struct ss{
	int node,nxt;
}e[600005];
inline int read()
{
	int re=0,k=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')k=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){re=(re<<1)+(re<<3)+ch-48;ch=getchar();}
	return re*k;
}
int u,v,n,col[200005],f[200005],del[200005],s[200005],sz[200005],head[200005],tot,g[200005];
void add(int u,int v)
{
	e[++tot].nxt=head[u];
	e[tot].node=v;
	head[u]=tot;
}
void getsz(int x,int fa)
{
	for(int i=head[x];i;i=e[i].nxt)
	{
		int t=e[i].node;
		if(t==fa)continue;
		getsz(t,x);
		sz[x]+=sz[t];
	}
}
void dfs(int x,int fa)
{
	int lst=s[col[fa]];
	s[col[fa]]=x;
	if(s[col[x]])del[s[col[x]]]+=sz[x];
	else g[col[x]]+=sz[x];
	for(int i=head[x];i;i=e[i].nxt)
	{
		int t=e[i].node;
		if(t==fa)continue;
		dfs(t,x);
	}
	int now=sz[x]-del[x];
	//cerr<<col[fa]<<" "<<now<<" "<<sz[x]<<endl;
	f[col[fa]]+=now*(now+1)>>1;
	s[col[fa]]=lst;
}
signed main()
{
	n=read();
	for(int i=1;i<=n;i++)
	{
		col[i]=read();sz[i]=1;
	}
	for(int i=1;i<=n-1;i++)
	{
		u=read();
		v=read();
		add(u,v);
		add(v,u);
	}
	getsz(1,0);
	dfs(1,0);
	for(int i=1;i<=n;i++)
		g[i]=(n-g[i])*(n-g[i]+1)>>1;
	int ans=n*(n+1)>>1;
	for(int i=1;i<=n;i++)
	{
		printf("%lld\n",ans-f[i]-g[i]);
	}
}
### 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<bits/stdc++.h>
using namespace std;
#define int long long 
struct ss{
	int node,nxt;
}e[600005];
inline int read()
{
	int re=0,k=1;char ch=getchar();
	while(ch<'0'||ch>'9'){if(ch=='-')k=-1;ch=getchar();}
	while(ch>='0'&&ch<='9'){re=(re<<1)+(re<<3)+ch-48;ch=getchar();}
	return re*k;
}
int u,v,n,col[200005],f[200005],del[200005],s[200005],sz[200005],head[200005],tot,g[200005];
void add(int u,int v)
{
	e[++tot].nxt=head[u];
	e[tot].node=v;
	head[u]=tot;
}
void getsz(int x,int fa)
{
	for(int i=head[x];i;i=e[i].nxt)
	{
		int t=e[i].node;
		if(t==fa)continue;
		getsz(t,x);
		sz[x]+=sz[t];
	}
}
void dfs(int x,int fa)
{
	int lst=s[col[fa]];
	s[col[fa]]=x;
	if(s[col[x]])del[s[col[x]]]+=sz[x];
	else g[col[x]]+=sz[x];
	for(int i=head[x];i;i=e[i].nxt)
	{
		int t=e[i].node;
		if(t==fa)continue;
		dfs(t,x);
	}
	int now=sz[x]-del[x];
	//cerr<<col[fa]<<" "<<now<<" "<<sz[x]<<endl;
	f[col[fa]]+=now*(now+1)>>1;
	s[col[fa]]=lst;
}
signed main()
{
	n=read();
	for(int i=1;i<=n;i++)
	{
		col[i]=read();sz[i]=1;
	}
	for(int i=1;i<=n-1;i++)
	{
		u=read();
		v=read();
		add(u,v);
		add(v,u);
	}
	getsz(1,0);
	dfs(1,0);
	for(int i=1;i<=n;i++)
		g[i]=(n-g[i])*(n-g[i]+1)>>1;
	int ans=n*(n+1)>>1;
	for(int i=1;i<=n;i++)
	{
		printf("%lld\n",ans-f[i]-g[i]);
	}
}
```