#include<bits/stdc++.h>
using namespace std;
typedef long long i64;
const int N=2e5+5;

i64 ans[N];
int x[N];
int s[N];
int c[N];
vector<int>e[N];

void dfs(int v,int p){
	int i,l=x[c[v]];s[v]=1;
	for(i=0;i<e[v].size();i++)
		if(e[v][i]!=p)
			x[c[v]]=0,dfs(e[v][i],v),s[v]+=s[e[v][i]],
			ans[c[v]]-=1ll*(s[e[v][i]]-x[c[v]])*(s[e[v][i]]-x[c[v]]+1)/2;
	x[c[v]]=l+s[v];
}

int main()
{
	int n,i,u,v;
	scanf("%d",&n);
	for(i=1;i<=n;i++){
		scanf("%d",&c[i]);
		ans[c[i]]=1ll*n*(n+1)/2;
	}
	for(i=1;i<n;i++)
		scanf("%d%d",&u,&v),
		e[u].push_back(v),
		e[v].push_back(u);
	dfs(1,0);
	for(i=1;i<=n;i++)
		if(ans[i])ans[i]-=1ll*(n-x[i])*(n-x[i]+1)/2;
	for(i=1;i<=n;i++)
		printf("%lld\n",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>
using namespace std;
typedef long long i64;
const int N=2e5+5;

i64 ans[N];
int x[N];
int s[N];
int c[N];
vector<int>e[N];

void dfs(int v,int p){
	int i,l=x[c[v]];s[v]=1;
	for(i=0;i<e[v].size();i++)
		if(e[v][i]!=p)
			x[c[v]]=0,dfs(e[v][i],v),s[v]+=s[e[v][i]],
			ans[c[v]]-=1ll*(s[e[v][i]]-x[c[v]])*(s[e[v][i]]-x[c[v]]+1)/2;
	x[c[v]]=l+s[v];
}

int main()
{
	int n,i,u,v;
	scanf("%d",&n);
	for(i=1;i<=n;i++){
		scanf("%d",&c[i]);
		ans[c[i]]=1ll*n*(n+1)/2;
	}
	for(i=1;i<n;i++)
		scanf("%d%d",&u,&v),
		e[u].push_back(v),
		e[v].push_back(u);
	dfs(1,0);
	for(i=1;i<=n;i++)
		if(ans[i])ans[i]-=1ll*(n-x[i])*(n-x[i]+1)/2;
	for(i=1;i<=n;i++)
		printf("%lld\n",ans[i]);
	return 0;
}
```