#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N = 2e5 + 10;
#define fi first
#define se second
#define pb push_back
#define mp make_pair
int n,a[N];
vector<int>v[N];
int sz[N],s[N];
LL get(int x){
	return 1ll*x*(x-1)/2;
}
LL ans[N];
void dfs(int x,int y){
	sz[x]=1;int pre=s[a[x]];
	for(auto k:v[x])if(k!=y){
		int g=s[a[x]];
		dfs(k,x);
		ans[a[x]]-=get(sz[k]-s[a[x]]+g);  
		sz[x]+=sz[k];
	}
	s[a[x]]=pre+sz[x];
	return ;
}

int main() {
  ios::sync_with_stdio(false);
  cin>>n;
  for(int i=1;i<=n;i++)cin>>a[i],ans[a[i]]++;
  for(int i=1;i<n;i++){
  	int s,t;
  	cin>>s>>t;
  	v[s].pb(t);
  	v[t].pb(s);
  }
  for(int i=1;i<=n;i++){
  	ans[i]+=1ll*n*(n-1)/2;
  }
  dfs(1,0);
  for(int i=1;i<=n;i++){
  	ans[i]-=get(n-s[i]);
  	cout<<ans[i]<<'\n';
  }
 	return 0;
}
### Prompt

Create a solution in CPP for 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;
typedef long long LL;
const int N = 2e5 + 10;
#define fi first
#define se second
#define pb push_back
#define mp make_pair
int n,a[N];
vector<int>v[N];
int sz[N],s[N];
LL get(int x){
	return 1ll*x*(x-1)/2;
}
LL ans[N];
void dfs(int x,int y){
	sz[x]=1;int pre=s[a[x]];
	for(auto k:v[x])if(k!=y){
		int g=s[a[x]];
		dfs(k,x);
		ans[a[x]]-=get(sz[k]-s[a[x]]+g);  
		sz[x]+=sz[k];
	}
	s[a[x]]=pre+sz[x];
	return ;
}

int main() {
  ios::sync_with_stdio(false);
  cin>>n;
  for(int i=1;i<=n;i++)cin>>a[i],ans[a[i]]++;
  for(int i=1;i<n;i++){
  	int s,t;
  	cin>>s>>t;
  	v[s].pb(t);
  	v[t].pb(s);
  }
  for(int i=1;i<=n;i++){
  	ans[i]+=1ll*n*(n-1)/2;
  }
  dfs(1,0);
  for(int i=1;i<=n;i++){
  	ans[i]-=get(n-s[i]);
  	cout<<ans[i]<<'\n';
  }
 	return 0;
}
```