        #include <bits/stdc++.h>
        using namespace std;
   #define int long long
const int MAX = 510000;
const int MOD =  1000000007;
int sz[200000],zen[200000],c[200000],ans[200000];bool used[200000];
vector<int> hen[200000];
int dfssz(int a){
     used[a]=true;
     int s=1;
     for(auto&& i : hen[a])if(!used[i])s+=dfssz(i);
     sz[a]=s;
     return s;
}
void dfs(int a){
     used[a]=true;
     zen[c[a]]-=sz[a];
     ans[c[a]]+=(zen[c[a]]+1)*sz[a];
     int tmp=zen[c[a]],x=sz[a]-1;
     for(auto&& i : hen[a])if(!used[i]){
          x-=sz[i];
          ans[c[a]]+=x*sz[i];
          zen[c[a]]=sz[i];
          dfs(i);
     }
     zen[c[a]]=tmp;
     return;
}

     signed main(){
       int n;
       cin>>n;
      for(int i=0;i<n;i++){
           cin>>c[i];c[i]--;
      }
      int a,b;
      for(int i=0;i<n-1;i++){
           cin>>a>>b;a--;b--;
           hen[a].push_back(b);
           hen[b].push_back(a);
      }
      fill(ans,ans+n,0);
      fill(used,used+n,false);
      fill(zen,zen+n,n);
      dfssz(0);
       fill(used,used+n,false);
      dfs(0);
      for(int i=0;i<n;i++)cout<<ans[i]<<endl;

      }

        

### Prompt

Please create a solution in CPP 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 <bits/stdc++.h>
        using namespace std;
   #define int long long
const int MAX = 510000;
const int MOD =  1000000007;
int sz[200000],zen[200000],c[200000],ans[200000];bool used[200000];
vector<int> hen[200000];
int dfssz(int a){
     used[a]=true;
     int s=1;
     for(auto&& i : hen[a])if(!used[i])s+=dfssz(i);
     sz[a]=s;
     return s;
}
void dfs(int a){
     used[a]=true;
     zen[c[a]]-=sz[a];
     ans[c[a]]+=(zen[c[a]]+1)*sz[a];
     int tmp=zen[c[a]],x=sz[a]-1;
     for(auto&& i : hen[a])if(!used[i]){
          x-=sz[i];
          ans[c[a]]+=x*sz[i];
          zen[c[a]]=sz[i];
          dfs(i);
     }
     zen[c[a]]=tmp;
     return;
}

     signed main(){
       int n;
       cin>>n;
      for(int i=0;i<n;i++){
           cin>>c[i];c[i]--;
      }
      int a,b;
      for(int i=0;i<n-1;i++){
           cin>>a>>b;a--;b--;
           hen[a].push_back(b);
           hen[b].push_back(a);
      }
      fill(ans,ans+n,0);
      fill(used,used+n,false);
      fill(zen,zen+n,n);
      dfssz(0);
       fill(used,used+n,false);
      dfs(0);
      for(int i=0;i<n;i++)cout<<ans[i]<<endl;

      }

        

```