#include<bits/stdc++.h>
#define ll long long
using namespace std;
int n,tot,sz[200010],a[200010],e[500010],rt[500010],hd[500010],nt[500010];
ll f[500010];
map<int,int>mp[200010];
void build(int x,int y){
	tot++;
	e[tot]=y;
	nt[tot]=hd[x];hd[x]=tot;
}
void hb(int &x,int y){
	int t,k;
	if(mp[x].size()<mp[y].size())swap(x,y);
	for(map<int,int>::iterator it=mp[y].begin();it!=mp[y].end();it++){
		t=it->first;k=it->second;
		mp[x][t]+=k;
	}
	mp[y].clear();
}
void dfs(int x,int fa){
	int i,w;
	sz[x]=1;rt[x]=x;
	for(i=hd[x];i;i=nt[i]){
		if(e[i]==fa)continue;
		dfs(e[i],x);
		sz[x]+=sz[e[i]];
		hb(rt[x],rt[e[i]]);
	}
	mp[rt[x]][a[x]]=sz[x];
	if(x==1){
		for(i=1;i<=n;i++){
			w=mp[rt[x]][i];
			f[i]+=1ll*(n-w)*(n-w+1)/2;
		}
	}
	 else{
	 	w=mp[rt[x]][a[fa]];
	 	f[a[fa]]+=1ll*(sz[x]-w)*(sz[x]-w+1)/2;
	 }
}
int main(){
	int i,x,y;
	/*scanf("%d",&n);
	for(i=1;i<=n;i++)*/
	//mp[1][1]=mp[1][2]=0;
	//printf("%d",mp[1].size());
	scanf("%d",&n);
	for(i=1;i<=n;i++)scanf("%d",&a[i]);
	for(i=1;i<n;i++){
		scanf("%d%d",&x,&y);
		build(x,y);build(y,x);
	}
	dfs(1,0);
	for(i=1;i<=n;i++)printf("%lld\n",1ll*n*(n+1)/2-f[i]);
}
### Prompt

Please create a solution in cpp 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>
#define ll long long
using namespace std;
int n,tot,sz[200010],a[200010],e[500010],rt[500010],hd[500010],nt[500010];
ll f[500010];
map<int,int>mp[200010];
void build(int x,int y){
	tot++;
	e[tot]=y;
	nt[tot]=hd[x];hd[x]=tot;
}
void hb(int &x,int y){
	int t,k;
	if(mp[x].size()<mp[y].size())swap(x,y);
	for(map<int,int>::iterator it=mp[y].begin();it!=mp[y].end();it++){
		t=it->first;k=it->second;
		mp[x][t]+=k;
	}
	mp[y].clear();
}
void dfs(int x,int fa){
	int i,w;
	sz[x]=1;rt[x]=x;
	for(i=hd[x];i;i=nt[i]){
		if(e[i]==fa)continue;
		dfs(e[i],x);
		sz[x]+=sz[e[i]];
		hb(rt[x],rt[e[i]]);
	}
	mp[rt[x]][a[x]]=sz[x];
	if(x==1){
		for(i=1;i<=n;i++){
			w=mp[rt[x]][i];
			f[i]+=1ll*(n-w)*(n-w+1)/2;
		}
	}
	 else{
	 	w=mp[rt[x]][a[fa]];
	 	f[a[fa]]+=1ll*(sz[x]-w)*(sz[x]-w+1)/2;
	 }
}
int main(){
	int i,x,y;
	/*scanf("%d",&n);
	for(i=1;i<=n;i++)*/
	//mp[1][1]=mp[1][2]=0;
	//printf("%d",mp[1].size());
	scanf("%d",&n);
	for(i=1;i<=n;i++)scanf("%d",&a[i]);
	for(i=1;i<n;i++){
		scanf("%d%d",&x,&y);
		build(x,y);build(y,x);
	}
	dfs(1,0);
	for(i=1;i<=n;i++)printf("%lld\n",1ll*n*(n+1)/2-f[i]);
}
```