#include <bits/stdc++.h>
#define mk make_pair
#define fs first
#define sc second
using namespace std;
typedef long long ll;
typedef long double ld;

#define MAX 200010
#define COL 200010

vector <int> a[MAX];

int sz[MAX];
void sz_dfs(int x, int papa){
  sz[x]=1;
  int nxt;
  for(int i=0; i<a[x].size(); ++i){
    nxt=a[x][i];
    if(nxt!=papa){
      sz_dfs(nxt, x);
      sz[x]+=sz[nxt];
    }
  }
}
int lst[200010], cnt[200010], col[200010];
ll ans[200010], n, nn[200010];
void dfs(int x, int pa){
	int tmp=lst[col[x]];
	lst[col[x]]=x;
	vector<ll> v;
	ll tmp1=0;
	for(int i=0; i<a[x].size(); ++i){
		if(a[x][i]==pa){
			continue;
		}
		dfs(a[x][i], x);
		v.push_back(sz[a[x][i]]-cnt[x]);
		tmp1+=sz[a[x][i]]-cnt[x];
		cnt[x]=0;
	}
	ll tmp2=tmp1;
	v.push_back(nn[col[x]]-tmp1-1);
	tmp1=nn[col[x]];
	for(auto it:v){
		// cout<<x<<" "<<col[x]<<" "<<it<<endl;
		ans[col[x]]+=it*(tmp1-it);
	}
	ans[col[x]]+=tmp1+1;
	lst[col[x]]=tmp;
	cnt[tmp]+=sz[x];
	nn[col[x]]-=tmp2+1;
}
int main(){
  int x, y;
  cin>>n;
  for(int i=1; i<=n; ++i){
    scanf("%d",&col[i]);
    nn[i]=n;
  }
  for(int i=0; i<n-1; ++i){
    scanf("%d%d",&x,&y);
    a[x].push_back(y);
    a[y].push_back(x);
  }
  sz_dfs(1,0);
  dfs(1,0);
  for(int i=1; i<=n; ++i){
  	ans[i]/=2;
    printf("%lld\n",ans[i]);
  }
  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 mk make_pair
#define fs first
#define sc second
using namespace std;
typedef long long ll;
typedef long double ld;

#define MAX 200010
#define COL 200010

vector <int> a[MAX];

int sz[MAX];
void sz_dfs(int x, int papa){
  sz[x]=1;
  int nxt;
  for(int i=0; i<a[x].size(); ++i){
    nxt=a[x][i];
    if(nxt!=papa){
      sz_dfs(nxt, x);
      sz[x]+=sz[nxt];
    }
  }
}
int lst[200010], cnt[200010], col[200010];
ll ans[200010], n, nn[200010];
void dfs(int x, int pa){
	int tmp=lst[col[x]];
	lst[col[x]]=x;
	vector<ll> v;
	ll tmp1=0;
	for(int i=0; i<a[x].size(); ++i){
		if(a[x][i]==pa){
			continue;
		}
		dfs(a[x][i], x);
		v.push_back(sz[a[x][i]]-cnt[x]);
		tmp1+=sz[a[x][i]]-cnt[x];
		cnt[x]=0;
	}
	ll tmp2=tmp1;
	v.push_back(nn[col[x]]-tmp1-1);
	tmp1=nn[col[x]];
	for(auto it:v){
		// cout<<x<<" "<<col[x]<<" "<<it<<endl;
		ans[col[x]]+=it*(tmp1-it);
	}
	ans[col[x]]+=tmp1+1;
	lst[col[x]]=tmp;
	cnt[tmp]+=sz[x];
	nn[col[x]]-=tmp2+1;
}
int main(){
  int x, y;
  cin>>n;
  for(int i=1; i<=n; ++i){
    scanf("%d",&col[i]);
    nn[i]=n;
  }
  for(int i=0; i<n-1; ++i){
    scanf("%d%d",&x,&y);
    a[x].push_back(y);
    a[y].push_back(x);
  }
  sz_dfs(1,0);
  dfs(1,0);
  for(int i=1; i<=n; ++i){
  	ans[i]/=2;
    printf("%lld\n",ans[i]);
  }
  return 0;
}

```