#include<bits/stdc++.h>
using namespace std;
#define pfi(a) printf("%d",a);
#define pfl(a) printf("%lld",a);
#define pfin(a) printf("%d\n",a);
#define pfln(a) printf("%lld\n",a);
#define sfi(a) scanf("%d",&a);
#define sfl(a) scanf("%lld",&a);
#define pb push_back
#define f(i,a,b) for(int i=a;i<b;i++)
#define ll long long
#define endl "\n"
const int MAXN=3e5+5;
const int M= 1e9+7;

int col[MAXN];

ll res[MAXN],sub[MAXN],val[MAXN];

vector<int> v[MAXN];

void dfs(int node,int par){
	sub[node]=1;

	for(int x:v[node]){
		if(x==par) continue;

		dfs(x,node);
		sub[node]+=sub[x];
	}
}

void solve(int node,int par){
	ll N=val[col[node]];

	ll fin=(N-sub[node]+1)*(sub[node]);

	ll sum=0;

	for(int x:v[node]){
		if(x==par) continue;
		fin+=(sum*sub[x]);
		sum+=sub[x];
	}

	res[col[node]]+=fin;

	for(int x:v[node]){
		if(x==par) continue;
		val[col[node]]=sub[x];
		solve(x,node);
	}

	val[col[node]]=N-sub[node];

}

int main(){	
	
	int n;
	sfi(n)

	f(i,1,n+1){
		sfi(col[i])
		val[i]=n;
	}

	f(i,0,n-1){
		int x,y;
		sfi(x) sfi(y)

		v[x].push_back(y);
		v[y].push_back(x);
	}

	dfs(1,1);

	solve(1,1);

	for(int i=1;i<=n;i++){
		cout<<res[i]<<"\n";
	}


	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;
#define pfi(a) printf("%d",a);
#define pfl(a) printf("%lld",a);
#define pfin(a) printf("%d\n",a);
#define pfln(a) printf("%lld\n",a);
#define sfi(a) scanf("%d",&a);
#define sfl(a) scanf("%lld",&a);
#define pb push_back
#define f(i,a,b) for(int i=a;i<b;i++)
#define ll long long
#define endl "\n"
const int MAXN=3e5+5;
const int M= 1e9+7;

int col[MAXN];

ll res[MAXN],sub[MAXN],val[MAXN];

vector<int> v[MAXN];

void dfs(int node,int par){
	sub[node]=1;

	for(int x:v[node]){
		if(x==par) continue;

		dfs(x,node);
		sub[node]+=sub[x];
	}
}

void solve(int node,int par){
	ll N=val[col[node]];

	ll fin=(N-sub[node]+1)*(sub[node]);

	ll sum=0;

	for(int x:v[node]){
		if(x==par) continue;
		fin+=(sum*sub[x]);
		sum+=sub[x];
	}

	res[col[node]]+=fin;

	for(int x:v[node]){
		if(x==par) continue;
		val[col[node]]=sub[x];
		solve(x,node);
	}

	val[col[node]]=N-sub[node];

}

int main(){	
	
	int n;
	sfi(n)

	f(i,1,n+1){
		sfi(col[i])
		val[i]=n;
	}

	f(i,0,n-1){
		int x,y;
		sfi(x) sfi(y)

		v[x].push_back(y);
		v[y].push_back(x);
	}

	dfs(1,1);

	solve(1,1);

	for(int i=1;i<=n;i++){
		cout<<res[i]<<"\n";
	}


	return 0;	
}
```