#include<bits/stdc++.h>
using namespace std;
typedef long long int ll;
typedef long double ld;
typedef pair<ll,ll> p2;
ll c[200005],sz[200005],tin[200005],ans[200005],timer=1;
vector<vector<ll>> res(200005);
ll func(ll n){
	return n*(n+1)/2;
}
void dfs(ll i,ll p,vector<vector<ll>> &v){
	sz[i]=1;
	tin[i]=timer++;
	for(ll j=0;j<ll(v[i].size());j++)
	if(v[i][j]!=p){
		dfs(v[i][j],i,v);
		sz[i]+=sz[v[i][j]];
		ll val=sz[v[i][j]];
		while(!res[c[i]].empty()&&tin[res[c[i]].back()]>=tin[i]){
			val-=sz[res[c[i]].back()];
			res[c[i]].pop_back();
		}
		ans[c[i]]+=func(val);
	}
	res[c[i]].push_back(i);
}
void solve(){
	ll n;
	cin>>n;
	for(ll i=1;i<=n;i++)
	cin>>c[i];
	vector<vector<ll>> v(n+1);
	for(ll i=0;i<n-1;i++){
		ll a,b;
		cin>>a>>b;
		v[a].push_back(b);
		v[b].push_back(a);
	}
	dfs(1,1,v);
	for(ll i=1;i<=n;i++){
		ll sum=n;
		for(ll j=0;j<ll(res[i].size());j++)
		sum-=sz[res[i][j]];
		ans[i]+=func(sum);
	}
	for(ll i=1;i<=n;i++){
		ans[i]=func(n)-ans[i];
		cout<<ans[i]<<"\n";
	}
}
int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    ll qc=1;
    for(ll i=1;i<=qc;i++)
    solve();
}

### Prompt

Develop a solution in Cpp 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;
typedef long long int ll;
typedef long double ld;
typedef pair<ll,ll> p2;
ll c[200005],sz[200005],tin[200005],ans[200005],timer=1;
vector<vector<ll>> res(200005);
ll func(ll n){
	return n*(n+1)/2;
}
void dfs(ll i,ll p,vector<vector<ll>> &v){
	sz[i]=1;
	tin[i]=timer++;
	for(ll j=0;j<ll(v[i].size());j++)
	if(v[i][j]!=p){
		dfs(v[i][j],i,v);
		sz[i]+=sz[v[i][j]];
		ll val=sz[v[i][j]];
		while(!res[c[i]].empty()&&tin[res[c[i]].back()]>=tin[i]){
			val-=sz[res[c[i]].back()];
			res[c[i]].pop_back();
		}
		ans[c[i]]+=func(val);
	}
	res[c[i]].push_back(i);
}
void solve(){
	ll n;
	cin>>n;
	for(ll i=1;i<=n;i++)
	cin>>c[i];
	vector<vector<ll>> v(n+1);
	for(ll i=0;i<n-1;i++){
		ll a,b;
		cin>>a>>b;
		v[a].push_back(b);
		v[b].push_back(a);
	}
	dfs(1,1,v);
	for(ll i=1;i<=n;i++){
		ll sum=n;
		for(ll j=0;j<ll(res[i].size());j++)
		sum-=sz[res[i][j]];
		ans[i]+=func(sum);
	}
	for(ll i=1;i<=n;i++){
		ans[i]=func(n)-ans[i];
		cout<<ans[i]<<"\n";
	}
}
int main(){
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    cout.tie(NULL);
    ll qc=1;
    for(ll i=1;i<=qc;i++)
    solve();
}

```