#include <bits/stdc++.h>

#define int long long

using namespace std;

const int oo = 2e18;

const int MAXN = 2e5 + 100;

#ifndef ONLINE_JUDGE 
#include "/media/brayand/7801-B00D/debugger.h"
#else
template<typename T>
void db(T x){}
#endif

int N;

int arr[MAXN];

vector<int> L[MAXN];

map<int,int> mp[MAXN];

int sz[MAXN];

int res[MAXN];

void DFS(int nodo, int padre)
{
	sz[nodo] = 1;

	for(auto v : L[nodo])
	{
		if(v != padre)
		{
			DFS(v,nodo);

			int value = sz[v]-mp[v][arr[nodo]];

			res[arr[nodo]] += value*(value+1)/2;

			sz[nodo] += sz[v];
			if(mp[nodo].size() < mp[v].size())swap(mp[nodo],mp[v]);
			for(auto x : mp[v])
			{
				mp[nodo][x.first] += x.second;
			}
		}
	}

	mp[nodo][arr[nodo]] = sz[nodo];

}

int32_t main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

    cin >> N;

    for(int i = 1 ; i <= N ; i++)
    {
    	cin >> arr[i];
    }

    for(int i = 1 ; i <= N-1 ; i++)
    {
    	int a, b;
    	cin >> a >> b;
    	L[a].push_back(b);
    	L[b].push_back(a);
    }

    DFS(1,-1);

    for(int i = 1 ; i <= N ; i++)
    {
    	int value = sz[1]-mp[1][i];
    	res[i] += value*(value+1)/2;
    }

    for(int i = 1 ; i <= N ; i++)
    {
    	cout << N*(N+1)/2-res[i] << '\n';
    }

    return 0;
}
### Prompt

Construct a Cpp code solution to the problem outlined:

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 oo = 2e18;

const int MAXN = 2e5 + 100;

#ifndef ONLINE_JUDGE 
#include "/media/brayand/7801-B00D/debugger.h"
#else
template<typename T>
void db(T x){}
#endif

int N;

int arr[MAXN];

vector<int> L[MAXN];

map<int,int> mp[MAXN];

int sz[MAXN];

int res[MAXN];

void DFS(int nodo, int padre)
{
	sz[nodo] = 1;

	for(auto v : L[nodo])
	{
		if(v != padre)
		{
			DFS(v,nodo);

			int value = sz[v]-mp[v][arr[nodo]];

			res[arr[nodo]] += value*(value+1)/2;

			sz[nodo] += sz[v];
			if(mp[nodo].size() < mp[v].size())swap(mp[nodo],mp[v]);
			for(auto x : mp[v])
			{
				mp[nodo][x.first] += x.second;
			}
		}
	}

	mp[nodo][arr[nodo]] = sz[nodo];

}

int32_t main()
{
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);

    cin >> N;

    for(int i = 1 ; i <= N ; i++)
    {
    	cin >> arr[i];
    }

    for(int i = 1 ; i <= N-1 ; i++)
    {
    	int a, b;
    	cin >> a >> b;
    	L[a].push_back(b);
    	L[b].push_back(a);
    }

    DFS(1,-1);

    for(int i = 1 ; i <= N ; i++)
    {
    	int value = sz[1]-mp[1][i];
    	res[i] += value*(value+1)/2;
    }

    for(int i = 1 ; i <= N ; i++)
    {
    	cout << N*(N+1)/2-res[i] << '\n';
    }

    return 0;
}
```