#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=200005;
int tot,color[maxn],head[maxn],sz[maxn],sum[maxn];
ll ans[maxn];
struct node{
	int to,next;
}edge[maxn*2];
void add_edge(int u,int v){
	++tot;
	edge[tot].to=v;
	edge[tot].next=head[u];
	head[u]=tot;
}
ll f(ll n){return n*(n+1)/2;}
void dfs(int u,int fa){	
	int c=color[u],last,delta,v;
	sz[u]=1;
	ll pre=sum[c];
	for(int i=head[u];i;i=edge[i].next){
		v=edge[i].to;
		if(v==fa) continue;
		last=sum[c];
		dfs(v,u);
		delta=sum[c]-last;
		sz[u]+=sz[v];
		ans[c]-=f(sz[v]-delta);
	}
	sum[c]=pre+sz[u];
}
int main(){
	int n;
	scanf("%d",&n);
	for(int i=1;i<=n;i++){
		scanf("%d",&color[i]);
		ans[i]=f(n);
	}
	for(int i=1;i<=n-1;i++){
		int u,v;
		scanf("%d%d",&u,&v);
		add_edge(u,v);
		add_edge(v,u);
	}
	dfs(1,0);
	for(int i=1;i<=n;i++){
		ans[i]-=f(n-sum[i]);
		printf("%lld\n",ans[i]);
	}
	return 0;
}
### Prompt

Please formulate 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
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn=200005;
int tot,color[maxn],head[maxn],sz[maxn],sum[maxn];
ll ans[maxn];
struct node{
	int to,next;
}edge[maxn*2];
void add_edge(int u,int v){
	++tot;
	edge[tot].to=v;
	edge[tot].next=head[u];
	head[u]=tot;
}
ll f(ll n){return n*(n+1)/2;}
void dfs(int u,int fa){	
	int c=color[u],last,delta,v;
	sz[u]=1;
	ll pre=sum[c];
	for(int i=head[u];i;i=edge[i].next){
		v=edge[i].to;
		if(v==fa) continue;
		last=sum[c];
		dfs(v,u);
		delta=sum[c]-last;
		sz[u]+=sz[v];
		ans[c]-=f(sz[v]-delta);
	}
	sum[c]=pre+sz[u];
}
int main(){
	int n;
	scanf("%d",&n);
	for(int i=1;i<=n;i++){
		scanf("%d",&color[i]);
		ans[i]=f(n);
	}
	for(int i=1;i<=n-1;i++){
		int u,v;
		scanf("%d%d",&u,&v);
		add_edge(u,v);
		add_edge(v,u);
	}
	dfs(1,0);
	for(int i=1;i<=n;i++){
		ans[i]-=f(n-sum[i]);
		printf("%lld\n",ans[i]);
	}
	return 0;
}
```