#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<ll,ll> pii;
ll read(){
	ll a=0,b=getchar(),c=1;
	while(!isdigit(b))c=b=='-'?-1:1,b=getchar();
	while(isdigit(b))a=a*10+b-'0',b=getchar();
	return a*c;
}
int a[200005];
ll n,ans[200005],b[200005],c[200005];
vector<int> g[200005];
ll gg(ll x){
	return x*(x-1)/2+x;
}
void dfs(int x,int y){
	c[x]=1;
	ll d=a[x],s=b[d];
	for(int i=0;i<g[x].size();i++)
		if(g[x][i]!=y){
			int tmp=b[d];
			dfs(g[x][i],x);
			c[x]+=c[g[x][i]];
			ans[d]-=gg(c[g[x][i]]-(b[d]-tmp));
		}
	b[d]=s+c[x];
}
int main(){
	n=read();
	for(int i=1;i<=n;i++)
		a[i]=read();
	for(int i=1;i<n;i++){
		int x=read(),y=read();
		g[x].push_back(y);
		g[y].push_back(x);
	}
	for(int i=1;i<=n;i++)
		ans[i]=gg(n);
	dfs(1,0);
	for(int i=1;i<=n;i++){
		ans[i]-=gg(n-b[i]);
		printf("%lld\n",ans[i]);
	}
	return 0;
}
### Prompt

Construct a cpp code solution to the problem outlined:

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;
typedef pair<ll,ll> pii;
ll read(){
	ll a=0,b=getchar(),c=1;
	while(!isdigit(b))c=b=='-'?-1:1,b=getchar();
	while(isdigit(b))a=a*10+b-'0',b=getchar();
	return a*c;
}
int a[200005];
ll n,ans[200005],b[200005],c[200005];
vector<int> g[200005];
ll gg(ll x){
	return x*(x-1)/2+x;
}
void dfs(int x,int y){
	c[x]=1;
	ll d=a[x],s=b[d];
	for(int i=0;i<g[x].size();i++)
		if(g[x][i]!=y){
			int tmp=b[d];
			dfs(g[x][i],x);
			c[x]+=c[g[x][i]];
			ans[d]-=gg(c[g[x][i]]-(b[d]-tmp));
		}
	b[d]=s+c[x];
}
int main(){
	n=read();
	for(int i=1;i<=n;i++)
		a[i]=read();
	for(int i=1;i<n;i++){
		int x=read(),y=read();
		g[x].push_back(y);
		g[y].push_back(x);
	}
	for(int i=1;i<=n;i++)
		ans[i]=gg(n);
	dfs(1,0);
	for(int i=1;i<=n;i++){
		ans[i]-=gg(n-b[i]);
		printf("%lld\n",ans[i]);
	}
	return 0;
}
```