#include<bits/stdc++.h>
using namespace std;
int c[200005];
vector<long long> G[200005];
long long path_num[200005];
long long ch_num[200005];
long long n;
long long ans;
long long dfs(long long u,long long v){
	long long pre = ch_num[c[u]];
	long long cnt_ch_num = 1;
	long long update = ch_num[c[u]];
	for(auto i:G[u]){
		if(i!=v){
			long long cntt = dfs(i,u);
			cnt_ch_num += cntt;
			long long delta = ch_num[c[u]] - update;
			long long k = cntt - delta;
			path_num[c[u]] += (k+1)*k/2;
			update = ch_num[c[u]];
		}
	}
	
	
	ch_num[c[u]] = pre + cnt_ch_num;
	return cnt_ch_num;
}
int main(){
	//freopen("in.txt","r",stdin);
	cin>>n;
	for(long long i=0;i<n;i++) {
		cin>>c[i];
		c[i]--;
	}
	
	long long t1,t2;
	for(long long i=0;i<n-1;i++){
		cin>>t1>>t2;
		t1--;t2--;
		G[t1].push_back(t2);
		G[t2].push_back(t1);
	}
	
	dfs(0,-1);
	for(long long i=0;i<n;i++){
		
			long long k=n-ch_num[i];
			ans = (n+1)*n/2 - path_num[i] - (k+1)*k/2;
		
		cout<<ans<<endl;
	}
	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;
int c[200005];
vector<long long> G[200005];
long long path_num[200005];
long long ch_num[200005];
long long n;
long long ans;
long long dfs(long long u,long long v){
	long long pre = ch_num[c[u]];
	long long cnt_ch_num = 1;
	long long update = ch_num[c[u]];
	for(auto i:G[u]){
		if(i!=v){
			long long cntt = dfs(i,u);
			cnt_ch_num += cntt;
			long long delta = ch_num[c[u]] - update;
			long long k = cntt - delta;
			path_num[c[u]] += (k+1)*k/2;
			update = ch_num[c[u]];
		}
	}
	
	
	ch_num[c[u]] = pre + cnt_ch_num;
	return cnt_ch_num;
}
int main(){
	//freopen("in.txt","r",stdin);
	cin>>n;
	for(long long i=0;i<n;i++) {
		cin>>c[i];
		c[i]--;
	}
	
	long long t1,t2;
	for(long long i=0;i<n-1;i++){
		cin>>t1>>t2;
		t1--;t2--;
		G[t1].push_back(t2);
		G[t2].push_back(t1);
	}
	
	dfs(0,-1);
	for(long long i=0;i<n;i++){
		
			long long k=n-ch_num[i];
			ans = (n+1)*n/2 - path_num[i] - (k+1)*k/2;
		
		cout<<ans<<endl;
	}
	return 0;
}
```