#pragma GCC optimize("Ofast")
#include<bits/stdc++.h>
#define int long long
#define gmax(x,y) x=max(x,y)
#define gmin(x,y) x=min(x,y)
#define F first
#define S second
#define P pair
#define FOR(i,a,b) for(int i=a;i<=b;i++)
#define rep(i,a,b) for(int i=a;i<b;i++)
#define V vector
#define RE return
#define ALL(a) a.begin(),a.end()
#define MP make_pair
#define PB emplace_back
#define PF emplace_front
#define FILL(a,b) memset(a,b,sizeof(a))
#define lwb lower_bound
#define upb upper_bound
using namespace std;
int c[200005],cnt[200005],ans[200005],sub[200005],n;
V<int> v[200005]; 
void get(int p,int fa){
	sub[p]=1;
	for(auto u:v[p])if(u!=fa){
		get(u,p);sub[p]+=sub[u];
	}
}
void dfs(int p,int fa){
	int pre=cnt[c[p]],t=n-sub[p]+1-pre,last=cnt[c[p]],sum=0,add=0;
	V<int> tv;
	for(auto u:v[p])if(u!=fa){
		dfs(u,p);
		ans[c[p]]-=(cnt[c[p]]-pre)*t;
		tv.PB(sub[u]-(cnt[c[p]]-pre));
		pre=cnt[c[p]];
	}
	reverse(ALL(tv));
	sum=sub[p]-1-(cnt[c[p]]-last);
	for(auto u:v[p])if(u!=fa){
		add+=tv.back()*(sum-tv.back());
		tv.pop_back();
	}
	ans[c[p]]=ans[c[p]]+sub[p]*t+add/2;
	cnt[c[p]]=last+sub[p];
}
signed main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	int x,y;
	cin>>n;
	FOR(i,1,n)cin>>c[i];
	rep(i,1,n){
		cin>>x>>y;
		v[x].PB(y);v[y].PB(x);
	}
	get(1,-1);
	dfs(1,-1);
	FOR(i,1,n)cout<<ans[i]<<'\n';
	RE 0;
}



### Prompt

Generate 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
#pragma GCC optimize("Ofast")
#include<bits/stdc++.h>
#define int long long
#define gmax(x,y) x=max(x,y)
#define gmin(x,y) x=min(x,y)
#define F first
#define S second
#define P pair
#define FOR(i,a,b) for(int i=a;i<=b;i++)
#define rep(i,a,b) for(int i=a;i<b;i++)
#define V vector
#define RE return
#define ALL(a) a.begin(),a.end()
#define MP make_pair
#define PB emplace_back
#define PF emplace_front
#define FILL(a,b) memset(a,b,sizeof(a))
#define lwb lower_bound
#define upb upper_bound
using namespace std;
int c[200005],cnt[200005],ans[200005],sub[200005],n;
V<int> v[200005]; 
void get(int p,int fa){
	sub[p]=1;
	for(auto u:v[p])if(u!=fa){
		get(u,p);sub[p]+=sub[u];
	}
}
void dfs(int p,int fa){
	int pre=cnt[c[p]],t=n-sub[p]+1-pre,last=cnt[c[p]],sum=0,add=0;
	V<int> tv;
	for(auto u:v[p])if(u!=fa){
		dfs(u,p);
		ans[c[p]]-=(cnt[c[p]]-pre)*t;
		tv.PB(sub[u]-(cnt[c[p]]-pre));
		pre=cnt[c[p]];
	}
	reverse(ALL(tv));
	sum=sub[p]-1-(cnt[c[p]]-last);
	for(auto u:v[p])if(u!=fa){
		add+=tv.back()*(sum-tv.back());
		tv.pop_back();
	}
	ans[c[p]]=ans[c[p]]+sub[p]*t+add/2;
	cnt[c[p]]=last+sub[p];
}
signed main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	int x,y;
	cin>>n;
	FOR(i,1,n)cin>>c[i];
	rep(i,1,n){
		cin>>x>>y;
		v[x].PB(y);v[y].PB(x);
	}
	get(1,-1);
	dfs(1,-1);
	FOR(i,1,n)cout<<ans[i]<<'\n';
	RE 0;
}



```