#include<bits/stdc++.h>
using namespace std;
const int N=2e5+5;
typedef long long LL;
struct Edge{
	int next,to;
}edge[N*2];
int c[N],c1[N],fi[N],se,Size[N],Size1[N],n,Sizec[N];
LL ans[N];
inline void add_edge(int u,int v){edge[++se].next=fi[u],fi[u]=se,edge[se].to=v;}
inline void add(int u,int v){add_edge(u,v),add_edge(v,u);}
void dfs(int x,int fa){
	int w=(c[x]!=c[fa]?c[fa]:0),fa1=c1[w];
	c1[w]=x;Size[x]=1;
	for(int i=fi[x];i;i=edge[i].next){
		int v=edge[i].to;if(v==fa)continue;
		dfs(v,x);Size[x]+=Size[v];
	}
	if(c[x]==c[fa])return;
	ans[w]-=1ll*(Size[x]-Size1[x])*(Size[x]-Size1[x]+1)/2;
	if(c1[c[x]])Size1[c1[c[x]]]+=Size[x];
	else Sizec[c[x]]+=Size[x];
	c1[w]=fa1;
}
int main(){
	int u,v;scanf("%d",&n);
	for(int i=1;i<=n;i++)scanf("%d",c+i);
	for(int i=1;i<n;i++)scanf("%d%d",&u,&v),add(u,v);
	dfs(1,0);
	for(int i=1;i<=n;i++)Sizec[i]=n-Sizec[i],ans[i]+=1ll*n*(n+1)/2-1ll*Sizec[i]*(Sizec[i]+1)/2,printf("%lld\n",ans[i]);
	return 0;
} 
### Prompt

Construct a Cpp code solution to the problem outlined:

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=2e5+5;
typedef long long LL;
struct Edge{
	int next,to;
}edge[N*2];
int c[N],c1[N],fi[N],se,Size[N],Size1[N],n,Sizec[N];
LL ans[N];
inline void add_edge(int u,int v){edge[++se].next=fi[u],fi[u]=se,edge[se].to=v;}
inline void add(int u,int v){add_edge(u,v),add_edge(v,u);}
void dfs(int x,int fa){
	int w=(c[x]!=c[fa]?c[fa]:0),fa1=c1[w];
	c1[w]=x;Size[x]=1;
	for(int i=fi[x];i;i=edge[i].next){
		int v=edge[i].to;if(v==fa)continue;
		dfs(v,x);Size[x]+=Size[v];
	}
	if(c[x]==c[fa])return;
	ans[w]-=1ll*(Size[x]-Size1[x])*(Size[x]-Size1[x]+1)/2;
	if(c1[c[x]])Size1[c1[c[x]]]+=Size[x];
	else Sizec[c[x]]+=Size[x];
	c1[w]=fa1;
}
int main(){
	int u,v;scanf("%d",&n);
	for(int i=1;i<=n;i++)scanf("%d",c+i);
	for(int i=1;i<n;i++)scanf("%d%d",&u,&v),add(u,v);
	dfs(1,0);
	for(int i=1;i<=n;i++)Sizec[i]=n-Sizec[i],ans[i]+=1ll*n*(n+1)/2-1ll*Sizec[i]*(Sizec[i]+1)/2,printf("%lld\n",ans[i]);
	return 0;
} 
```