#include<bits/stdc++.h>
using namespace std;
#define sqr(x) 1ll*x*(x+1)>>1ll
typedef long long ll;
int n,col[200100],sz[200100];
map<int,int>f[200100];
vector<int>v[200100],res[200100];
void dfs(int x,int fa){
	sz[x]=1;
	for(int y:v[x]){
		if(y==fa)continue;
		dfs(y,x),sz[x]+=sz[y];
		int cc=sz[y];
		if(f[y].find(col[x])!=f[y].end())cc-=f[y][col[x]];
		res[col[x]].push_back(cc);
		if(f[x].size()<f[y].size())swap(f[x],f[y]);
		for(auto i:f[y])f[x][i.first]+=i.second; 
	}
	f[x][col[x]]=sz[x];
}
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++)scanf("%d",&col[i]);
	for(int i=1,x,y;i<n;i++)scanf("%d%d",&x,&y),v[x].push_back(y),v[y].push_back(x);
	dfs(1,0);
	for(int i=1;i<=n;i++)res[i].push_back(n-f[1][i]);
	for(int i=1;i<=n;i++){
		ll ans=sqr(n);
		for(auto j:res[i])ans-=sqr(j);
		printf("%lld\n",ans);
	}
	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;
#define sqr(x) 1ll*x*(x+1)>>1ll
typedef long long ll;
int n,col[200100],sz[200100];
map<int,int>f[200100];
vector<int>v[200100],res[200100];
void dfs(int x,int fa){
	sz[x]=1;
	for(int y:v[x]){
		if(y==fa)continue;
		dfs(y,x),sz[x]+=sz[y];
		int cc=sz[y];
		if(f[y].find(col[x])!=f[y].end())cc-=f[y][col[x]];
		res[col[x]].push_back(cc);
		if(f[x].size()<f[y].size())swap(f[x],f[y]);
		for(auto i:f[y])f[x][i.first]+=i.second; 
	}
	f[x][col[x]]=sz[x];
}
int main(){
	scanf("%d",&n);
	for(int i=1;i<=n;i++)scanf("%d",&col[i]);
	for(int i=1,x,y;i<n;i++)scanf("%d%d",&x,&y),v[x].push_back(y),v[y].push_back(x);
	dfs(1,0);
	for(int i=1;i<=n;i++)res[i].push_back(n-f[1][i]);
	for(int i=1;i<=n;i++){
		ll ans=sqr(n);
		for(auto j:res[i])ans-=sqr(j);
		printf("%lld\n",ans);
	}
	return 0;
}
```