#include <bits/stdc++.h>

#define rep(i,n) for(int i=0;i<n;i++)
#define repn(i,n) for(int i=1;i<=n;i++)
#define LL long long
#define pii pair <int,int>
#define fi first
#define se second
#define pb push_back
#define mpr make_pair

using namespace std;

const LL MOD=1e9+7;

LL n,c[200010],x,y,sonn[200010],ans[200010],lst[200010],a[200010],fa[200010];
vector <LL> g[200010],v[200010];

LL dfs_pre(LL pos,LL par)
{
	fa[pos]=par;
	sonn[pos]=1;
	rep(i,g[pos].size()) if(g[pos][i]!=par) sonn[pos]+=dfs_pre(g[pos][i],pos);
	return sonn[pos];
}

void dfs(LL pos,LL par)
{
	if(lst[c[pos]]==0) a[c[pos]]+=sonn[pos];
	else v[lst[c[pos]]].pb(pos);
	LL sv=lst[c[pos]];
	rep(i,g[pos].size()) if(g[pos][i]!=par)
	{
		lst[c[pos]]=g[pos][i];
		dfs(g[pos][i],pos);
	}
	lst[c[pos]]=sv;
}

int main()
{
	cin>>n;
	repn(i,n) scanf("%lld",&c[i]);
	rep(i,n-1)
	{
		scanf("%lld%lld",&x,&y);
		g[x].pb(y);
		g[y].pb(x);
	}
	dfs_pre(1,0);
	dfs(1,0);
	for(int i=2;i<=n;i++)
	{
		LL sum=0;
		rep(j,v[i].size()) sum+=sonn[v[i][j]];
		//cout<<i<<' '<<sum<<' '<<c[fa[i]]<<endl;
		ans[c[fa[i]]]+=(sonn[i]-sum)*(sonn[i]-sum+1)/2;
	}
	//repn(i,n) cout<<ans[i]<<' ';cout<<endl;
	repn(i,n) ans[i]+=(n-a[i])*(n-a[i]+1)/2;
	repn(i,n) printf("%lld\n",n*(n+1)/2-ans[i]);
	return 0;
}
### 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 rep(i,n) for(int i=0;i<n;i++)
#define repn(i,n) for(int i=1;i<=n;i++)
#define LL long long
#define pii pair <int,int>
#define fi first
#define se second
#define pb push_back
#define mpr make_pair

using namespace std;

const LL MOD=1e9+7;

LL n,c[200010],x,y,sonn[200010],ans[200010],lst[200010],a[200010],fa[200010];
vector <LL> g[200010],v[200010];

LL dfs_pre(LL pos,LL par)
{
	fa[pos]=par;
	sonn[pos]=1;
	rep(i,g[pos].size()) if(g[pos][i]!=par) sonn[pos]+=dfs_pre(g[pos][i],pos);
	return sonn[pos];
}

void dfs(LL pos,LL par)
{
	if(lst[c[pos]]==0) a[c[pos]]+=sonn[pos];
	else v[lst[c[pos]]].pb(pos);
	LL sv=lst[c[pos]];
	rep(i,g[pos].size()) if(g[pos][i]!=par)
	{
		lst[c[pos]]=g[pos][i];
		dfs(g[pos][i],pos);
	}
	lst[c[pos]]=sv;
}

int main()
{
	cin>>n;
	repn(i,n) scanf("%lld",&c[i]);
	rep(i,n-1)
	{
		scanf("%lld%lld",&x,&y);
		g[x].pb(y);
		g[y].pb(x);
	}
	dfs_pre(1,0);
	dfs(1,0);
	for(int i=2;i<=n;i++)
	{
		LL sum=0;
		rep(j,v[i].size()) sum+=sonn[v[i][j]];
		//cout<<i<<' '<<sum<<' '<<c[fa[i]]<<endl;
		ans[c[fa[i]]]+=(sonn[i]-sum)*(sonn[i]-sum+1)/2;
	}
	//repn(i,n) cout<<ans[i]<<' ';cout<<endl;
	repn(i,n) ans[i]+=(n-a[i])*(n-a[i]+1)/2;
	repn(i,n) printf("%lld\n",n*(n+1)/2-ans[i]);
	return 0;
}
```