#include<bits/stdc++.h>
#define ll long long 
#define inf 0x7fffffff
#define clr(c) memset(c,0,sizeof(c))
using namespace std;
int mod=1e9+7;
const int M=1e6+10;
const int N=2e5+10;
inline ll read(){
    ll b=1,sum=0; char c=getchar();
    while(!isdigit(c)){if(c=='-') b=-1; c=getchar();}
    while(isdigit(c)){sum=sum*10+c-'0'; c=getchar();}
    return b*sum;
}
int n,m,k,T;
ll ans[N];
int col[N];
int vis[N],sz[N],s[N];
bool vv[N];
vector<vector<int> > e;
ll f(int x) {return 1ll*x*(x+1)/2;}
void dfs(int x,int _fa) {
	int i,j;
	sz[x]=1;
	ll res=0,temp=vis[col[x]];
//	vis[col[x]]=0;
	for(auto y:e[x]) {
		if(y==_fa) continue;
		vis[col[x]]=0;
		dfs(y,x);
		res+=vis[col[x]];
		sz[x]+=sz[y];
		s[col[x]]-=vis[col[x]];
		ans[col[x]]+=f(sz[y]-vis[col[x]]);
	}
	vis[col[x]]=sz[x]+temp;
	s[col[x]]+=sz[x];
}
int main() 
{
    int i,j;
	n=read();
	e.resize(n+1);
	for(i=1;i<=n;i++) {
		col[i]=read();
		vv[col[i]]=1;
	}

	for(i=1;i<=n-1;i++) {
		int x=read(),y=read();
		e[x].push_back(y);
		e[y].push_back(x);
	}
	dfs(1,0);
	for(i=1;i<=n;i++) {
		printf("%lld\n",f(n)-f(n-vis[i])-ans[i]);
	}
    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
#include<bits/stdc++.h>
#define ll long long 
#define inf 0x7fffffff
#define clr(c) memset(c,0,sizeof(c))
using namespace std;
int mod=1e9+7;
const int M=1e6+10;
const int N=2e5+10;
inline ll read(){
    ll b=1,sum=0; char c=getchar();
    while(!isdigit(c)){if(c=='-') b=-1; c=getchar();}
    while(isdigit(c)){sum=sum*10+c-'0'; c=getchar();}
    return b*sum;
}
int n,m,k,T;
ll ans[N];
int col[N];
int vis[N],sz[N],s[N];
bool vv[N];
vector<vector<int> > e;
ll f(int x) {return 1ll*x*(x+1)/2;}
void dfs(int x,int _fa) {
	int i,j;
	sz[x]=1;
	ll res=0,temp=vis[col[x]];
//	vis[col[x]]=0;
	for(auto y:e[x]) {
		if(y==_fa) continue;
		vis[col[x]]=0;
		dfs(y,x);
		res+=vis[col[x]];
		sz[x]+=sz[y];
		s[col[x]]-=vis[col[x]];
		ans[col[x]]+=f(sz[y]-vis[col[x]]);
	}
	vis[col[x]]=sz[x]+temp;
	s[col[x]]+=sz[x];
}
int main() 
{
    int i,j;
	n=read();
	e.resize(n+1);
	for(i=1;i<=n;i++) {
		col[i]=read();
		vv[col[i]]=1;
	}

	for(i=1;i<=n-1;i++) {
		int x=read(),y=read();
		e[x].push_back(y);
		e[y].push_back(x);
	}
	dfs(1,0);
	for(i=1;i<=n;i++) {
		printf("%lld\n",f(n)-f(n-vis[i])-ans[i]);
	}
    return 0;
}

```