#include <bits/stdc++.h>
using namespace std;

#define ll long long
#define mod 1000000007
#define pb push_back
#define pi pair<int,int>
#define vi vector<int>
#define vpi vector< pi > 
#define IOS ios_base::sync_with_stdio(false),cin.tie(NULL)
const int N=200003;
vi v[N];
ll c[N],a[N],val[N],res[N],sub[N];
int dfs(int x,int p)
{
    for(int y:v[x])
        if(y!=p)
            sub[x]+=dfs(y,x);
    sub[x]+=1;
    return sub[x];
}
void solv(int x,int p)
{
    ll n=val[c[x]];
    ll add=(n-sub[x]+1)*sub[x],sm=0;
    for(int y:v[x])
    {
        if(y==p) continue;
        add+=1ll*sub[y]*sm;
        sm+=sub[y];
    }
    res[c[x]]+=add;
    //val[c[x]]=sub[x]-1;
    for(int y:v[x])
    {
        if(y==p) continue;
        val[c[x]]=sub[y];
        solv(y,x);
    }
    val[c[x]]=n-sub[x];
}
int main()
{
    int n,x,y;
    cin>>n;
    for(int i=1;i<=n;i++)
    {
        cin>>c[i];
        val[i]=n;
    }
    for(int i=1;i<n;i++)
    {
        cin>>x>>y;
        v[x].pb(y);
        v[y].pb(x);
    }
    dfs(1,0);
    solv(1,0);
    for(int i=1;i<=n;i++)
    {
        cout<<res[i]<<'\n';
    }
    return 0;
}
### Prompt

Please provide a Cpp coded solution 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;

#define ll long long
#define mod 1000000007
#define pb push_back
#define pi pair<int,int>
#define vi vector<int>
#define vpi vector< pi > 
#define IOS ios_base::sync_with_stdio(false),cin.tie(NULL)
const int N=200003;
vi v[N];
ll c[N],a[N],val[N],res[N],sub[N];
int dfs(int x,int p)
{
    for(int y:v[x])
        if(y!=p)
            sub[x]+=dfs(y,x);
    sub[x]+=1;
    return sub[x];
}
void solv(int x,int p)
{
    ll n=val[c[x]];
    ll add=(n-sub[x]+1)*sub[x],sm=0;
    for(int y:v[x])
    {
        if(y==p) continue;
        add+=1ll*sub[y]*sm;
        sm+=sub[y];
    }
    res[c[x]]+=add;
    //val[c[x]]=sub[x]-1;
    for(int y:v[x])
    {
        if(y==p) continue;
        val[c[x]]=sub[y];
        solv(y,x);
    }
    val[c[x]]=n-sub[x];
}
int main()
{
    int n,x,y;
    cin>>n;
    for(int i=1;i<=n;i++)
    {
        cin>>c[i];
        val[i]=n;
    }
    for(int i=1;i<n;i++)
    {
        cin>>x>>y;
        v[x].pb(y);
        v[y].pb(x);
    }
    dfs(1,0);
    solv(1,0);
    for(int i=1;i<=n;i++)
    {
        cout<<res[i]<<'\n';
    }
    return 0;
}
```