#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define FOR(i,a,b) for(ll i=(a);i<(b);i++)
#define REP(i,a) FOR(i,0,a)
const ll MAX_N=2e5;
ll N,c[MAX_N];
vector<ll> G[MAX_N];
ll tsz[MAX_N],tszadd;
ll ans[MAX_N];
ll cntp(ll n){
	return n+n*(n-1)/2;
}
void dfs(ll v,ll p){
	ll tszb=tsz[c[v]]+tszadd;
	for(auto e:G[v]){
		if(e!=p){
			ll tmp=tsz[c[v]]+tszadd;
			dfs(e,v);
			ans[c[v]]-=cntp(tsz[c[v]]+tszadd-tmp);
		}
	}
	tszadd++;
	tsz[c[v]]=tszb-tszadd;
}
int main(){
	cin.tie(0);
	cin.sync_with_stdio(false);
	cin>>N;
	REP(i,N)cin>>c[i];
	REP(i,N)c[i]--;
	REP(i,N-1){
		ll a,b;
		cin>>a>>b;
		a--;
		b--;
		G[a].push_back(b);
		G[b].push_back(a);
	}
	REP(i,N){
		ans[i]=cntp(N);
	}
	dfs(0,-1);
	REP(k,N){
		ans[k]-=cntp(tsz[k]+tszadd);
	}
	REP(k,N){
		cout<<ans[k]<<endl;
	}
}

### 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 ll;
#define FOR(i,a,b) for(ll i=(a);i<(b);i++)
#define REP(i,a) FOR(i,0,a)
const ll MAX_N=2e5;
ll N,c[MAX_N];
vector<ll> G[MAX_N];
ll tsz[MAX_N],tszadd;
ll ans[MAX_N];
ll cntp(ll n){
	return n+n*(n-1)/2;
}
void dfs(ll v,ll p){
	ll tszb=tsz[c[v]]+tszadd;
	for(auto e:G[v]){
		if(e!=p){
			ll tmp=tsz[c[v]]+tszadd;
			dfs(e,v);
			ans[c[v]]-=cntp(tsz[c[v]]+tszadd-tmp);
		}
	}
	tszadd++;
	tsz[c[v]]=tszb-tszadd;
}
int main(){
	cin.tie(0);
	cin.sync_with_stdio(false);
	cin>>N;
	REP(i,N)cin>>c[i];
	REP(i,N)c[i]--;
	REP(i,N-1){
		ll a,b;
		cin>>a>>b;
		a--;
		b--;
		G[a].push_back(b);
		G[b].push_back(a);
	}
	REP(i,N){
		ans[i]=cntp(N);
	}
	dfs(0,-1);
	REP(k,N){
		ans[k]-=cntp(tsz[k]+tszadd);
	}
	REP(k,N){
		cout<<ans[k]<<endl;
	}
}

```