#include <iostream>
#include <cstdio>
#include <cstdlib>
using namespace std;
const int maxn=1e6+10;
int n,c[maxn],st[maxn],to[maxn],nt[maxn],topt,si[maxn],num[maxn];
long long ans[maxn];
inline void add(int x,int y){to[++topt]=y; nt[topt]=st[x]; st[x]=topt;}
inline long long f(int x){return 1ll*x*(x+1)/2;}
void dfs(int x,int fa)
{
	si[x]=1; int k=num[c[x]],p=st[x];
	while (p)
	{
		if (to[p]!=fa) 
		{
			int now=num[c[x]]; dfs(to[p],x); 
			si[x]+=si[to[p]]; now=si[to[p]]-(num[c[x]]-now); ans[c[x]]-=f(now);
		}
		p=nt[p];
	}
	num[c[x]]=k+si[x];
}
int main()
{
	scanf("%d",&n);
	for (int i=1;i<=n;i++) scanf("%d",&c[i]),ans[i]=f(n);
	for (int i=1;i<n;i++) {int x,y; scanf("%d%d",&x,&y); add(x,y); add(y,x);}
	dfs(1,0);
	for (int i=1;i<=n;i++) ans[i]-=f(n-num[i]),printf("%lld\n",ans[i]);
return 0;
}
### Prompt

Generate a CPP solution to 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 <iostream>
#include <cstdio>
#include <cstdlib>
using namespace std;
const int maxn=1e6+10;
int n,c[maxn],st[maxn],to[maxn],nt[maxn],topt,si[maxn],num[maxn];
long long ans[maxn];
inline void add(int x,int y){to[++topt]=y; nt[topt]=st[x]; st[x]=topt;}
inline long long f(int x){return 1ll*x*(x+1)/2;}
void dfs(int x,int fa)
{
	si[x]=1; int k=num[c[x]],p=st[x];
	while (p)
	{
		if (to[p]!=fa) 
		{
			int now=num[c[x]]; dfs(to[p],x); 
			si[x]+=si[to[p]]; now=si[to[p]]-(num[c[x]]-now); ans[c[x]]-=f(now);
		}
		p=nt[p];
	}
	num[c[x]]=k+si[x];
}
int main()
{
	scanf("%d",&n);
	for (int i=1;i<=n;i++) scanf("%d",&c[i]),ans[i]=f(n);
	for (int i=1;i<n;i++) {int x,y; scanf("%d%d",&x,&y); add(x,y); add(y,x);}
	dfs(1,0);
	for (int i=1;i<=n;i++) ans[i]-=f(n-num[i]),printf("%lld\n",ans[i]);
return 0;
}
```