#include<bits/stdc++.h>
using namespace std;
#define FOR(i,a,b) for (int i=(a);i<(b);i++)
#define RFOR(i,a,b) for (int i=(b)-1;i>=(a);i--)
#define REP(i,n) for (int i=0;i<(n);i++)
#define RREP(i,n) for (int i=(n)-1;i>=0;i--)
typedef long long LL;
int N;
LL A[200001];
vector<int>G[200001];
LL cnt[200001];
LL ans[200001];


LL calc(LL x){
	LL ans=x*(x-1)/2;
	ans+=x;
	return max((LL)0,ans);
}

int dfs2(int x,int p){
	int size=1;
	LL t=0;
	REP(i,G[x].size()){
		if(G[x][i]!=p){
			LL now=cnt[A[x]];//開始時の数
			int s=dfs2(G[x][i],x);
			LL now2=cnt[A[x]];//終了時の値
			LL res=s-(now2-now);
			ans[A[x]]+=max(LL(0),calc(res));
			t+=res;
			size+=s;
		}
	}
	cnt[A[x]]+=(t+1);
	return size;
}

int main(){
	cin>>N;
	REP(i,N){
		cin>>A[i];
		A[i]--;
	}
	REP(i,N-1){
		int a,b;
		cin>>a>>b;
		a--;b--;
		G[a].push_back(b);
		G[b].push_back(a);
	}
	REP(i,N){
		cnt[i]=0;
		ans[i]=0;
	}
	dfs2(0,-1);
	REP(i,N){
		//cout<<"i"<<i<<" "<<ans[i]<<endl;
		cout<<calc(N)-ans[i]-calc(N-cnt[i])<<endl;
	}
	return 0;
}
### Prompt

Please provide a Cpp coded solution to the problem described below:

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 FOR(i,a,b) for (int i=(a);i<(b);i++)
#define RFOR(i,a,b) for (int i=(b)-1;i>=(a);i--)
#define REP(i,n) for (int i=0;i<(n);i++)
#define RREP(i,n) for (int i=(n)-1;i>=0;i--)
typedef long long LL;
int N;
LL A[200001];
vector<int>G[200001];
LL cnt[200001];
LL ans[200001];


LL calc(LL x){
	LL ans=x*(x-1)/2;
	ans+=x;
	return max((LL)0,ans);
}

int dfs2(int x,int p){
	int size=1;
	LL t=0;
	REP(i,G[x].size()){
		if(G[x][i]!=p){
			LL now=cnt[A[x]];//開始時の数
			int s=dfs2(G[x][i],x);
			LL now2=cnt[A[x]];//終了時の値
			LL res=s-(now2-now);
			ans[A[x]]+=max(LL(0),calc(res));
			t+=res;
			size+=s;
		}
	}
	cnt[A[x]]+=(t+1);
	return size;
}

int main(){
	cin>>N;
	REP(i,N){
		cin>>A[i];
		A[i]--;
	}
	REP(i,N-1){
		int a,b;
		cin>>a>>b;
		a--;b--;
		G[a].push_back(b);
		G[b].push_back(a);
	}
	REP(i,N){
		cnt[i]=0;
		ans[i]=0;
	}
	dfs2(0,-1);
	REP(i,N){
		//cout<<"i"<<i<<" "<<ans[i]<<endl;
		cout<<calc(N)-ans[i]-calc(N-cnt[i])<<endl;
	}
	return 0;
}
```