#include<bits/stdc++.h>
using namespace std;
const int N=2e5+10;
typedef long long ll;
int n,c[N],sz[N],all,now[N],u,v;
vector<int>e[N];
ll ans[N];
void dfs(int u,int fa){
	sz[u]=1;
	int tmp=now[c[u]];//进入u的子树前now[c[u]]的值 
	for(auto &v:e[u]){
		if(v==fa)continue;
		int pre=all-now[c[u]];
		dfs(v,u);
		sz[u]+=sz[v];
		int nex=all-now[c[u]];
		ans[c[u]]-=1ll*(nex-pre)*(nex-pre-1)/2;
	}
	all++;
	now[c[u]]=tmp+sz[u];//离开u的子树后now[c[u]]的值 
}
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;++i){
		scanf("%d",&c[i]);
		ans[i]=1ll*n*(n-1)/2;
	} 
	for(int i=1;i<n;++i){
		scanf("%d%d",&u,&v);
		e[u].push_back(v);
		e[v].push_back(u);
	}
	dfs(1,-1);
	for(int i=1;i<=n;++i){
		int pre=0,nex=all-now[i];
		ans[i]-=1ll*(nex-pre)*(nex-pre-1)/2;
	}
	for(int i=1;i<=n;++i){
		ans[c[i]]++;
	}
	for(int i=1;i<=n;++i){
		printf("%lld\n",ans[i]);
	}
	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<bits/stdc++.h>
using namespace std;
const int N=2e5+10;
typedef long long ll;
int n,c[N],sz[N],all,now[N],u,v;
vector<int>e[N];
ll ans[N];
void dfs(int u,int fa){
	sz[u]=1;
	int tmp=now[c[u]];//进入u的子树前now[c[u]]的值 
	for(auto &v:e[u]){
		if(v==fa)continue;
		int pre=all-now[c[u]];
		dfs(v,u);
		sz[u]+=sz[v];
		int nex=all-now[c[u]];
		ans[c[u]]-=1ll*(nex-pre)*(nex-pre-1)/2;
	}
	all++;
	now[c[u]]=tmp+sz[u];//离开u的子树后now[c[u]]的值 
}
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;++i){
		scanf("%d",&c[i]);
		ans[i]=1ll*n*(n-1)/2;
	} 
	for(int i=1;i<n;++i){
		scanf("%d%d",&u,&v);
		e[u].push_back(v);
		e[v].push_back(u);
	}
	dfs(1,-1);
	for(int i=1;i<=n;++i){
		int pre=0,nex=all-now[i];
		ans[i]-=1ll*(nex-pre)*(nex-pre-1)/2;
	}
	for(int i=1;i<=n;++i){
		ans[c[i]]++;
	}
	for(int i=1;i<=n;++i){
		printf("%lld\n",ans[i]);
	}
	return 0;
} 
```