#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int col[200010];
ll n,ans[200010],sum[200010],sz[200010];//sum[i]是指在一定范围内离根最近的有颜色i的点的子树大小（或子树的一部分） 
vector<int> g[200010];
ll mul(ll x) {
	return x*(x-1)/2+x;
}
void dfs(int pos,int fa) {
	sz[pos]=1;
	ll c=col[pos],s=sum[c];
	for(int i=0;i<g[pos].size();++i)
		if(g[pos][i]!=fa) {
			int tmp=sum[c];
			dfs(g[pos][i],pos);
			sz[pos]+=sz[g[pos][i]];
			ans[c]-=mul(sz[g[pos][i]]-(sum[c]-tmp));
		}
	sum[c]=s+sz[pos];
}
int main()
{
	scanf("%lld",&n);
	for(int i=1;i<=n;++i) scanf("%d",&col[i]);
	for(int i=1,x,y;i<n;++i) {
		scanf("%d%d",&x,&y);
		g[x].push_back(y);
		g[y].push_back(x);
	}
	for(int i=1;i<=n;++i) ans[i]=mul(n);
	dfs(1,0);
	for(int i=1;i<=n;++i) ans[i]-=mul(n-sum[i]),printf("%lld\n",ans[i]);
	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>
using namespace std;
typedef long long ll;
int col[200010];
ll n,ans[200010],sum[200010],sz[200010];//sum[i]是指在一定范围内离根最近的有颜色i的点的子树大小（或子树的一部分） 
vector<int> g[200010];
ll mul(ll x) {
	return x*(x-1)/2+x;
}
void dfs(int pos,int fa) {
	sz[pos]=1;
	ll c=col[pos],s=sum[c];
	for(int i=0;i<g[pos].size();++i)
		if(g[pos][i]!=fa) {
			int tmp=sum[c];
			dfs(g[pos][i],pos);
			sz[pos]+=sz[g[pos][i]];
			ans[c]-=mul(sz[g[pos][i]]-(sum[c]-tmp));
		}
	sum[c]=s+sz[pos];
}
int main()
{
	scanf("%lld",&n);
	for(int i=1;i<=n;++i) scanf("%d",&col[i]);
	for(int i=1,x,y;i<n;++i) {
		scanf("%d%d",&x,&y);
		g[x].push_back(y);
		g[y].push_back(x);
	}
	for(int i=1;i<=n;++i) ans[i]=mul(n);
	dfs(1,0);
	for(int i=1;i<=n;++i) ans[i]-=mul(n-sum[i]),printf("%lld\n",ans[i]);
	return 0;
}
```