/*
读题不规范，爆零两行泪。
数据不清空，爆零两行泪。
多测不读完，爆零两行泪。
边界不特判，爆零两行泪。
贪心不证明，爆零两行泪。
D P 顺序错，爆零两行泪。
大小少等号，爆零两行泪。
变量不统一，爆零两行泪。
越界不判断，爆零两行泪。
调试不注释，爆零两行泪。
溢出不 l l，爆零两行泪。
*/
#include<bits/stdc++.h>
using namespace std;
#define pb push_back
#define ppb pop_back
typedef long long ll;
const int N=200000;
int n;
int a[N+1];
vector<int> nei[N+1];
int sz[N+1],cc_sz[N+1],cc_sz1[N+1];
int fa[N+1];
vector<int> stk,stk_c[N+1];
void dfs(int x=1){
//	cout<<x<<"\n";
	sz[x]=1;
	stk_c[a[x]].pb(stk.size());
	stk.pb(x);
	for(int i=0;i<nei[x].size();i++){
		int y=nei[x][i];
		if(y==fa[x])continue;
		fa[y]=x;
		dfs(y);
		sz[x]+=sz[y];
	}
	stk.ppb();
	stk_c[a[x]].ppb();
	if(stk_c[a[x]].size()>1)cc_sz[stk[stk_c[a[x]].back()+1]]-=sz[x];
	else cc_sz1[a[x]]-=sz[x];
}
ll ans[N+1];
int main(){
	cin>>n;
	for(int i=1;i<=n;i++)cin>>a[i];
	for(int i=1;i<n;i++){
		int x,y;
		cin>>x>>y;
		nei[x].pb(y);nei[y].pb(x);
	}
	stk.pb(0);
	for(int i=1;i<=n;i++)stk_c[i].pb(0);
	dfs();
	for(int i=1;i<=n;i++)cc_sz[i]+=sz[i],cc_sz1[i]+=n;
//	for(int i=1;i<=n;i++)cout<<cc_sz[i]<<" ";puts("");
//	for(int i=1;i<=n;i++)cout<<cc_sz1[i]<<" ";puts("");
	for(int i=1;i<=n;i++)ans[i]=1ll*n*(n+1)/2;
	for(int i=2;i<=n;i++)ans[a[fa[i]]]-=1ll*cc_sz[i]*(cc_sz[i]+1)/2;
	for(int i=1;i<=n;i++)ans[i]-=1ll*cc_sz1[i]*(cc_sz1[i]+1)/2;
	for(int i=1;i<=n;i++)cout<<ans[i]<<"\n";
	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
/*
读题不规范，爆零两行泪。
数据不清空，爆零两行泪。
多测不读完，爆零两行泪。
边界不特判，爆零两行泪。
贪心不证明，爆零两行泪。
D P 顺序错，爆零两行泪。
大小少等号，爆零两行泪。
变量不统一，爆零两行泪。
越界不判断，爆零两行泪。
调试不注释，爆零两行泪。
溢出不 l l，爆零两行泪。
*/
#include<bits/stdc++.h>
using namespace std;
#define pb push_back
#define ppb pop_back
typedef long long ll;
const int N=200000;
int n;
int a[N+1];
vector<int> nei[N+1];
int sz[N+1],cc_sz[N+1],cc_sz1[N+1];
int fa[N+1];
vector<int> stk,stk_c[N+1];
void dfs(int x=1){
//	cout<<x<<"\n";
	sz[x]=1;
	stk_c[a[x]].pb(stk.size());
	stk.pb(x);
	for(int i=0;i<nei[x].size();i++){
		int y=nei[x][i];
		if(y==fa[x])continue;
		fa[y]=x;
		dfs(y);
		sz[x]+=sz[y];
	}
	stk.ppb();
	stk_c[a[x]].ppb();
	if(stk_c[a[x]].size()>1)cc_sz[stk[stk_c[a[x]].back()+1]]-=sz[x];
	else cc_sz1[a[x]]-=sz[x];
}
ll ans[N+1];
int main(){
	cin>>n;
	for(int i=1;i<=n;i++)cin>>a[i];
	for(int i=1;i<n;i++){
		int x,y;
		cin>>x>>y;
		nei[x].pb(y);nei[y].pb(x);
	}
	stk.pb(0);
	for(int i=1;i<=n;i++)stk_c[i].pb(0);
	dfs();
	for(int i=1;i<=n;i++)cc_sz[i]+=sz[i],cc_sz1[i]+=n;
//	for(int i=1;i<=n;i++)cout<<cc_sz[i]<<" ";puts("");
//	for(int i=1;i<=n;i++)cout<<cc_sz1[i]<<" ";puts("");
	for(int i=1;i<=n;i++)ans[i]=1ll*n*(n+1)/2;
	for(int i=2;i<=n;i++)ans[a[fa[i]]]-=1ll*cc_sz[i]*(cc_sz[i]+1)/2;
	for(int i=1;i<=n;i++)ans[i]-=1ll*cc_sz1[i]*(cc_sz1[i]+1)/2;
	for(int i=1;i<=n;i++)cout<<ans[i]<<"\n";
	return 0;
}
```