#include<bits/stdc++.h>

using namespace std;

#define int long long

const int N=3e5;

int n;
int color[N];
int f[N];
int sum[N];
vector<int> adj[N];
int cnt;

void dfs(int u,int pr){
	int tmp=f[color[u]]+cnt;
	cnt++;
	for(auto v:adj[u]){
		if(v!=pr){
			f[color[u]]=-cnt;
			dfs(v,u);
			int s=f[color[u]]+cnt;
			sum[color[u]]+=(s*(s+1))/2;
		}
	}
	f[color[u]]=tmp-cnt;
}

signed main(){
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	cin>>n;
	for(int i=1;i<=n;i++)
		cin>>color[i];
	for(int i=1;i<n;i++){
		int u,v;
		cin>>u>>v;
		adj[u].push_back(v);
		adj[v].push_back(u);
	}
	dfs(1,-1);
	for(int i=1;i<=n;i++){
		int s=f[i]+cnt;
		sum[i]+=(s*(s+1))/2;
		cout<<(n*(n+1))/2-sum[i]<<"\n";
	}
}
### Prompt

In Cpp, your task is to solve 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

const int N=3e5;

int n;
int color[N];
int f[N];
int sum[N];
vector<int> adj[N];
int cnt;

void dfs(int u,int pr){
	int tmp=f[color[u]]+cnt;
	cnt++;
	for(auto v:adj[u]){
		if(v!=pr){
			f[color[u]]=-cnt;
			dfs(v,u);
			int s=f[color[u]]+cnt;
			sum[color[u]]+=(s*(s+1))/2;
		}
	}
	f[color[u]]=tmp-cnt;
}

signed main(){
	ios_base::sync_with_stdio(false);
	cin.tie(0);
	cout.tie(0);
	cin>>n;
	for(int i=1;i<=n;i++)
		cin>>color[i];
	for(int i=1;i<n;i++){
		int u,v;
		cin>>u>>v;
		adj[u].push_back(v);
		adj[v].push_back(u);
	}
	dfs(1,-1);
	for(int i=1;i<=n;i++){
		int s=f[i]+cnt;
		sum[i]+=(s*(s+1))/2;
		cout<<(n*(n+1))/2-sum[i]<<"\n";
	}
}
```