#include <cstdio>
#include <algorithm>
using namespace std;

#define N 200005
#define int long long

int n,c[N],num[N],ans[N],C[N];
int tot,head[2*N],than[2*N],to[2*N]; 

inline int read(){
	int s=0,t=1;
	char c=getchar();
	while((c<'0'||c>'9')&&c!='-')c=getchar(),t=-1;
	if(c=='-')c=getchar(),t=-1;
	while(c>='0'&&c<='9')s=(s<<1)+(s<<3)+(c^48),c=getchar();
	return s*t;
}

void add(int x,int y){
	tot++;
	to[tot]=y;
	than[tot]=head[x];
	head[x]=tot;
}

void dfs(int now,int p=-1){
	int tmp=C[c[now]];
	for(int i=head[now];i;i=than[i]){
		int t=to[i];
		if(t==p)continue;
		C[c[now]]=0;
		dfs(t,now);
		num[now]+=num[t];
		ans[c[now]]-=(num[t]-C[c[now]])*(num[t]-C[c[now]]+1)/2; 
	}
	C[c[now]]=tmp+num[now];
}

signed main(void){
	n=read();
	for(int i=0;i<n;i++)
	c[i]=read(),c[i]--,num[i]=1;
	for(int i=1;i<n;i++){
		int x,y;
		x=read()-1;
		y=read()-1;
		add(x,y);
		add(y,x);
	}
	dfs(0);
	for(int i=0;i<n;i++){
		int f=n*(n+1)/2;
		int f1=(n-C[i])*(n-C[i]+1)/2;
		printf("%lld\n",ans[i]+f-f1);
	}
//	for(int i=0;i<n;i++)
//	printf("%d ",ans[i]); 
} 
### 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
#include <cstdio>
#include <algorithm>
using namespace std;

#define N 200005
#define int long long

int n,c[N],num[N],ans[N],C[N];
int tot,head[2*N],than[2*N],to[2*N]; 

inline int read(){
	int s=0,t=1;
	char c=getchar();
	while((c<'0'||c>'9')&&c!='-')c=getchar(),t=-1;
	if(c=='-')c=getchar(),t=-1;
	while(c>='0'&&c<='9')s=(s<<1)+(s<<3)+(c^48),c=getchar();
	return s*t;
}

void add(int x,int y){
	tot++;
	to[tot]=y;
	than[tot]=head[x];
	head[x]=tot;
}

void dfs(int now,int p=-1){
	int tmp=C[c[now]];
	for(int i=head[now];i;i=than[i]){
		int t=to[i];
		if(t==p)continue;
		C[c[now]]=0;
		dfs(t,now);
		num[now]+=num[t];
		ans[c[now]]-=(num[t]-C[c[now]])*(num[t]-C[c[now]]+1)/2; 
	}
	C[c[now]]=tmp+num[now];
}

signed main(void){
	n=read();
	for(int i=0;i<n;i++)
	c[i]=read(),c[i]--,num[i]=1;
	for(int i=1;i<n;i++){
		int x,y;
		x=read()-1;
		y=read()-1;
		add(x,y);
		add(y,x);
	}
	dfs(0);
	for(int i=0;i<n;i++){
		int f=n*(n+1)/2;
		int f1=(n-C[i])*(n-C[i]+1)/2;
		printf("%lld\n",ans[i]+f-f1);
	}
//	for(int i=0;i<n;i++)
//	printf("%d ",ans[i]); 
} 
```