#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#define pb push_back
using namespace std;
typedef long long ll;
const int N=2e5+10;
vector<int>E[N];
int c[N],u,v;
int n,sz[N],num[N];
ll ans[N];
ll C(ll x)
{
    return 1ll*x*(x-1)/2+x;
}
void dfs(int u,int fa)
{
    sz[u]=1;
    for(int v:E[u])
    {
        if(v==fa)continue;
        dfs(v,u);
        sz[u]+=sz[v];
    }
}
void dfs2(int u,int fa)
{
    int tmp=num[c[u]];
    for(int v:E[u])
    {
        if(v==fa)continue;
        num[c[u]]=sz[v];
        dfs2(v,u);
        ans[c[u]]-=C(num[c[u]]);
    }
    num[c[u]]=tmp-sz[u];
}
int main()
{
	scanf("%d",&n);
    for(int i=1;i<=n;++i)
    {
        E[i].clear();
        num[i]=sz[i]=0;
    }
    for(int i=1;i<=n;++i)
    {
        scanf("%d",&c[i]);
    }
    for(int i=1;i<n;++i)
    {
        scanf("%d%d",&u,&v);
        E[u].pb(v),E[v].pb(u);
    }
    dfs(1,-1);
    for(int i=1;i<=n;++i){
 	   num[i]=sz[1];
   	   ans[i]=C(n);
//   	   printf("%d:%lld\n",i,ans[i]);
    }
    dfs2(1,-1);
    for(int i=1;i<=n;++i){
    	ans[i]-=C(num[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<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#define pb push_back
using namespace std;
typedef long long ll;
const int N=2e5+10;
vector<int>E[N];
int c[N],u,v;
int n,sz[N],num[N];
ll ans[N];
ll C(ll x)
{
    return 1ll*x*(x-1)/2+x;
}
void dfs(int u,int fa)
{
    sz[u]=1;
    for(int v:E[u])
    {
        if(v==fa)continue;
        dfs(v,u);
        sz[u]+=sz[v];
    }
}
void dfs2(int u,int fa)
{
    int tmp=num[c[u]];
    for(int v:E[u])
    {
        if(v==fa)continue;
        num[c[u]]=sz[v];
        dfs2(v,u);
        ans[c[u]]-=C(num[c[u]]);
    }
    num[c[u]]=tmp-sz[u];
}
int main()
{
	scanf("%d",&n);
    for(int i=1;i<=n;++i)
    {
        E[i].clear();
        num[i]=sz[i]=0;
    }
    for(int i=1;i<=n;++i)
    {
        scanf("%d",&c[i]);
    }
    for(int i=1;i<n;++i)
    {
        scanf("%d%d",&u,&v);
        E[u].pb(v),E[v].pb(u);
    }
    dfs(1,-1);
    for(int i=1;i<=n;++i){
 	   num[i]=sz[1];
   	   ans[i]=C(n);
//   	   printf("%d:%lld\n",i,ans[i]);
    }
    dfs2(1,-1);
    for(int i=1;i<=n;++i){
    	ans[i]-=C(num[i]);
    	printf("%lld\n",ans[i]);
	}
    return 0;
}
```