#include<bits/stdc++.h>
#define int long long
using namespace std;
const int N=2e5+10;
int n;
int color_count[N],color[N],total[N],sz[N],tmp_count[N];
vector<int> vi[N],group[N];
 
void start(int u, int parent) 
{
    sz[u]=1;
    for(auto v: vi[u]) 
	if(v!=parent) 
	{
        start(v,u);
        sz[u]+=sz[v];
    }
} 
void dfs(int u, int parent) 
{
    int before=tmp_count[color[u]];
    int cur=tmp_count[color[u]];
    for(auto v: vi[u]) 
	{
		if(v!=parent) 
		{
        	dfs(v,u);
        	int inc=tmp_count[color[u]]-cur;
        	group[color[u]].push_back(sz[v]-inc);
        	total[color[u]]+=sz[v]-inc;
        	cur=tmp_count[color[u]];
    	}
	}
    tmp_count[color[u]]=before+sz[u];
} 
signed main()  
{
    ios_base::sync_with_stdio(false); 
	cin.tie(0); 
	cout.tie(0);
    cin >> n;
    for(int i=1;i<=n;i++)
	{
		cin>>color[i];
		color_count[color[i]]++;
	}
    for(int i=1;i<n;i++)
	{
        int u,v; 
		cin>>u>>v;
        vi[u].push_back(v); 
		vi[v].push_back(u);
    }
    start(1,1);
    dfs(1,1);
    for(int i=1;i<=n;i++) 
	{
		if(total[i]+color_count[i]<n)
        	group[i].push_back(n-total[i]-color_count[i]);
	}
    for(int i=1;i<=n;i++)
	{
        int ans=n*(n-1)/2+n;
        for(auto num: group[i]) 
		{
            ans-=(num*(num-1)/2);
            ans-=num;
        }
        cout<<ans<<endl;
    }
}
### Prompt

In cpp, your task is to solve 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 int long long
using namespace std;
const int N=2e5+10;
int n;
int color_count[N],color[N],total[N],sz[N],tmp_count[N];
vector<int> vi[N],group[N];
 
void start(int u, int parent) 
{
    sz[u]=1;
    for(auto v: vi[u]) 
	if(v!=parent) 
	{
        start(v,u);
        sz[u]+=sz[v];
    }
} 
void dfs(int u, int parent) 
{
    int before=tmp_count[color[u]];
    int cur=tmp_count[color[u]];
    for(auto v: vi[u]) 
	{
		if(v!=parent) 
		{
        	dfs(v,u);
        	int inc=tmp_count[color[u]]-cur;
        	group[color[u]].push_back(sz[v]-inc);
        	total[color[u]]+=sz[v]-inc;
        	cur=tmp_count[color[u]];
    	}
	}
    tmp_count[color[u]]=before+sz[u];
} 
signed main()  
{
    ios_base::sync_with_stdio(false); 
	cin.tie(0); 
	cout.tie(0);
    cin >> n;
    for(int i=1;i<=n;i++)
	{
		cin>>color[i];
		color_count[color[i]]++;
	}
    for(int i=1;i<n;i++)
	{
        int u,v; 
		cin>>u>>v;
        vi[u].push_back(v); 
		vi[v].push_back(u);
    }
    start(1,1);
    dfs(1,1);
    for(int i=1;i<=n;i++) 
	{
		if(total[i]+color_count[i]<n)
        	group[i].push_back(n-total[i]-color_count[i]);
	}
    for(int i=1;i<=n;i++)
	{
        int ans=n*(n-1)/2+n;
        for(auto num: group[i]) 
		{
            ans-=(num*(num-1)/2);
            ans-=num;
        }
        cout<<ans<<endl;
    }
}
```