#include<bits/stdc++.h>
#define GO ios::sync_with_stdio(0), cin.tie(0), cout.tie(0)
using namespace std;
typedef long long ll;
typedef vector<ll> vec;
const int N = 5e5 + 5;
const int M = 1e6 + 5;
const int Mod = 1e9 + 7;

int n, a[N];
vec G[N];
ll sum[N];
int sz[N];
ll ans[N];

ll f(ll x)
{
	return x * (x + 1) / 2;
}

void GetSz(int u, int p)
{
	sz[u] = 1;
	for (auto v : G[u])
	{
		if (v == p)continue;
		GetSz(v, u);
		sz[u] += sz[v];
	}
}

int Last[N];

void DFS(int u, int p)
{
	int color = a[u];
	int par = Last[color];
	sum[par] += sz[u];
	Last[color] = u;
	for (auto v : G[u])
	{
		if (v == p)continue;
		DFS(v, u);
		ans[color] += f(sz[v] - sum[u]);
		sum[u] = 0;
	}
	Last[color] = par;
}

int main()
{
	GO;
	cin >> n;
	for (int i = 1; i <= n; i++)
		cin >> a[i];
	for(int i = 1; i < n; i++)
	{
		int u, v;
		cin >> u >> v;
		G[u].push_back(v);
		G[v].push_back(u);
	}
	for (int i = 1; i <= n; i++)
		Last[i] = n + i;
	GetSz(1, 1);
	DFS(1, 1);
	for (int i = 1; i <= n; i++)
	{
		if (i == a[1])
			continue;
		ans[i] += f(n - sum[i + n]);
	}
	for (int i = 1; i <= n; i++)
		cout << (f(n) - ans[i] )  << '\n';
	return 0;
}
### Prompt

Develop a solution in Cpp to the problem described below:

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 GO ios::sync_with_stdio(0), cin.tie(0), cout.tie(0)
using namespace std;
typedef long long ll;
typedef vector<ll> vec;
const int N = 5e5 + 5;
const int M = 1e6 + 5;
const int Mod = 1e9 + 7;

int n, a[N];
vec G[N];
ll sum[N];
int sz[N];
ll ans[N];

ll f(ll x)
{
	return x * (x + 1) / 2;
}

void GetSz(int u, int p)
{
	sz[u] = 1;
	for (auto v : G[u])
	{
		if (v == p)continue;
		GetSz(v, u);
		sz[u] += sz[v];
	}
}

int Last[N];

void DFS(int u, int p)
{
	int color = a[u];
	int par = Last[color];
	sum[par] += sz[u];
	Last[color] = u;
	for (auto v : G[u])
	{
		if (v == p)continue;
		DFS(v, u);
		ans[color] += f(sz[v] - sum[u]);
		sum[u] = 0;
	}
	Last[color] = par;
}

int main()
{
	GO;
	cin >> n;
	for (int i = 1; i <= n; i++)
		cin >> a[i];
	for(int i = 1; i < n; i++)
	{
		int u, v;
		cin >> u >> v;
		G[u].push_back(v);
		G[v].push_back(u);
	}
	for (int i = 1; i <= n; i++)
		Last[i] = n + i;
	GetSz(1, 1);
	DFS(1, 1);
	for (int i = 1; i <= n; i++)
	{
		if (i == a[1])
			continue;
		ans[i] += f(n - sum[i + n]);
	}
	for (int i = 1; i <= n; i++)
		cout << (f(n) - ans[i] )  << '\n';
	return 0;
}
```