#include <bits/stdc++.h>
using namespace std;
#define mod 998244353
#define MOD 1000000007
#define inf 0x3f3f3f3f
#define linf 0x3f3f3f3f3f3f3f3fll
typedef long long ll;
typedef pair<int,int> pii;
typedef unsigned long long ull;
const int maxn=2e5+10;
int era[maxn],sz[maxn];
int n;
int col[maxn];
vector<int> g[maxn];
ll res[maxn];
void dfs(int x,int fa)
{
	sz[x]=0;
	for(int i=0;i<g[x].size();i++)
	{
		int to=g[x][i];
		if(to==fa)continue;
		int rec=era[col[x]];
		dfs(to,x);
		sz[x]+=sz[to];
		int lft=sz[to]-(era[col[x]]-rec);
	//	cout<<"comb "<<x<<" "<<col[x]<<" "<<lft<<" "<<era[col[x]]<<" "<<rec<<" "<<sz[x]<<endl;
		res[col[x]]+=lft*1ll*(lft+1)/2;
		era[col[x]]=rec+sz[to];
	}
	sz[x]++;
	era[col[x]]++;
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)scanf("%d",&col[i]);
	for(int i=1;i<n;i++)
	{
		int a,b;
		scanf("%d%d",&a,&b);
		g[a].push_back(b);
		g[b].push_back(a);
	}
	dfs(1,0);
	for(int i=1;i<=n;i++)
	{
		int lft=n-era[i];
	//	cout<<i<<" "<<era[i]<<" "<<lft<<endl;
		res[i]+=lft*1ll*(lft+1)/2;
	}
	for(int i=1;i<=n;i++)
	{
		printf("%lld\n",n*1ll*(n+1)/2-res[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>
using namespace std;
#define mod 998244353
#define MOD 1000000007
#define inf 0x3f3f3f3f
#define linf 0x3f3f3f3f3f3f3f3fll
typedef long long ll;
typedef pair<int,int> pii;
typedef unsigned long long ull;
const int maxn=2e5+10;
int era[maxn],sz[maxn];
int n;
int col[maxn];
vector<int> g[maxn];
ll res[maxn];
void dfs(int x,int fa)
{
	sz[x]=0;
	for(int i=0;i<g[x].size();i++)
	{
		int to=g[x][i];
		if(to==fa)continue;
		int rec=era[col[x]];
		dfs(to,x);
		sz[x]+=sz[to];
		int lft=sz[to]-(era[col[x]]-rec);
	//	cout<<"comb "<<x<<" "<<col[x]<<" "<<lft<<" "<<era[col[x]]<<" "<<rec<<" "<<sz[x]<<endl;
		res[col[x]]+=lft*1ll*(lft+1)/2;
		era[col[x]]=rec+sz[to];
	}
	sz[x]++;
	era[col[x]]++;
}
int main()
{
	scanf("%d",&n);
	for(int i=1;i<=n;i++)scanf("%d",&col[i]);
	for(int i=1;i<n;i++)
	{
		int a,b;
		scanf("%d%d",&a,&b);
		g[a].push_back(b);
		g[b].push_back(a);
	}
	dfs(1,0);
	for(int i=1;i<=n;i++)
	{
		int lft=n-era[i];
	//	cout<<i<<" "<<era[i]<<" "<<lft<<endl;
		res[i]+=lft*1ll*(lft+1)/2;
	}
	for(int i=1;i<=n;i++)
	{
		printf("%lld\n",n*1ll*(n+1)/2-res[i]);
	}
	return 0;
}

```