#include <map>
#include <cstdio>
#include <vector>
#include <cstring>
#define pii pair<int,int>
#define s1 first
#define s2 second
#define N 200001
using namespace std;struct edge{int v,next;edge(){}edge(int _v,int _next){v=_v;next=_next;}}e[N<<1];int head[N],sz;
void init(){memset(head,-1,sizeof(head));sz=0;}void insert(int u,int v){e[++sz]=edge(v,head[u]);head[u]=sz;}
map<int,int> dp[N];vector<int> cc[N];int siz[N],n,c[N];long long ans[N];
void dfs(int u,int f)
{
	for(int i=head[u];~i;i=e[i].next)
	{
		if(e[i].v==f)continue;dfs(e[i].v,u);siz[u]+=siz[e[i].v];int tmp=siz[e[i].v];
		if(dp[e[i].v].count(c[u]))tmp-=dp[e[i].v][c[u]];cc[c[u]].push_back(tmp);
		if(dp[u].size()<dp[e[i].v].size())swap(dp[u],dp[e[i].v]);
		for(map<int,int>::iterator it=dp[e[i].v].begin();it!=dp[e[i].v].end();++it)dp[u][it->s1]+=it->s2;
	}dp[u][c[u]]=++siz[u];
}
int main()
{
	init();scanf("%d",&n);for(int i=1;i<=n;++i)scanf("%d",c+i);for(int i=1,u,v;i<n;++i)scanf("%d%d",&u,&v),insert(u,v),insert(v,u);
	dfs(1,1);for(int i=1;i<=n;++i)ans[i]=1ll*n*(n+1)/2;for(int i=1;i<=n;++i)cc[i].push_back(n-dp[1][i]);
	for(int i=1;i<=n;++i)for(int j=0;j<cc[i].size();++j)ans[i]-=1ll*cc[i][j]*(cc[i][j]+1)/2;for(int i=1;i<=n;++i)printf("%lld\n",ans[i]);putchar(10);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 <map>
#include <cstdio>
#include <vector>
#include <cstring>
#define pii pair<int,int>
#define s1 first
#define s2 second
#define N 200001
using namespace std;struct edge{int v,next;edge(){}edge(int _v,int _next){v=_v;next=_next;}}e[N<<1];int head[N],sz;
void init(){memset(head,-1,sizeof(head));sz=0;}void insert(int u,int v){e[++sz]=edge(v,head[u]);head[u]=sz;}
map<int,int> dp[N];vector<int> cc[N];int siz[N],n,c[N];long long ans[N];
void dfs(int u,int f)
{
	for(int i=head[u];~i;i=e[i].next)
	{
		if(e[i].v==f)continue;dfs(e[i].v,u);siz[u]+=siz[e[i].v];int tmp=siz[e[i].v];
		if(dp[e[i].v].count(c[u]))tmp-=dp[e[i].v][c[u]];cc[c[u]].push_back(tmp);
		if(dp[u].size()<dp[e[i].v].size())swap(dp[u],dp[e[i].v]);
		for(map<int,int>::iterator it=dp[e[i].v].begin();it!=dp[e[i].v].end();++it)dp[u][it->s1]+=it->s2;
	}dp[u][c[u]]=++siz[u];
}
int main()
{
	init();scanf("%d",&n);for(int i=1;i<=n;++i)scanf("%d",c+i);for(int i=1,u,v;i<n;++i)scanf("%d%d",&u,&v),insert(u,v),insert(v,u);
	dfs(1,1);for(int i=1;i<=n;++i)ans[i]=1ll*n*(n+1)/2;for(int i=1;i<=n;++i)cc[i].push_back(n-dp[1][i]);
	for(int i=1;i<=n;++i)for(int j=0;j<cc[i].size();++j)ans[i]-=1ll*cc[i][j]*(cc[i][j]+1)/2;for(int i=1;i<=n;++i)printf("%lld\n",ans[i]);putchar(10);return 0;
}
```