#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
vector<ll> g[200005];
ll col[200005];
ll son[200005];
ll colson[200005];
ll ans[200005];
ll n;
ll f(ll n)
{
	return n*(n+1)/2;
}
void dfs(ll u,ll par)
{
	ll temp=colson[col[u]];
	for(ll i=0;i<g[u].size();i++)
	{
		ll v=g[u][i];
		if(v==par) continue;
		colson[col[u]]=0;
		dfs(v,u);
		son[u]+=son[v];
		ans[col[u]]-=f(son[v]-colson[col[u]]);
	}
	colson[col[u]]=temp+son[u];
}
int main(void)
{
	scanf("%lld",&n);
	for(ll i=1;i<=n;i++)
	{
		scanf("%lld",&col[i]);
		son[i]=1;
		ans[i]=f(n);
	}
	for(ll i=0;i<n-1;i++)
	{
		ll u,v;
		scanf("%lld%lld",&u,&v);
		g[u].push_back(v);
		g[v].push_back(u);
	}
	dfs(1,0);
	for(ll i=1;i<=n;i++)
	{
		printf("%lld\n",ans[i]-f(n-colson[i]));
	}
}
### Prompt

Please provide a cpp coded solution 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>
using namespace std;
typedef long long ll;
vector<ll> g[200005];
ll col[200005];
ll son[200005];
ll colson[200005];
ll ans[200005];
ll n;
ll f(ll n)
{
	return n*(n+1)/2;
}
void dfs(ll u,ll par)
{
	ll temp=colson[col[u]];
	for(ll i=0;i<g[u].size();i++)
	{
		ll v=g[u][i];
		if(v==par) continue;
		colson[col[u]]=0;
		dfs(v,u);
		son[u]+=son[v];
		ans[col[u]]-=f(son[v]-colson[col[u]]);
	}
	colson[col[u]]=temp+son[u];
}
int main(void)
{
	scanf("%lld",&n);
	for(ll i=1;i<=n;i++)
	{
		scanf("%lld",&col[i]);
		son[i]=1;
		ans[i]=f(n);
	}
	for(ll i=0;i<n-1;i++)
	{
		ll u,v;
		scanf("%lld%lld",&u,&v);
		g[u].push_back(v);
		g[v].push_back(u);
	}
	dfs(1,0);
	for(ll i=1;i<=n;i++)
	{
		printf("%lld\n",ans[i]-f(n-colson[i]));
	}
}
```