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

struct tree{
    size_t n,r,md;
    vector<vector<int>> e;
    vector<int> c,sz;
    vector<ll> ans,C;
    explicit tree(int n):n(n),e(n),c(n),C(n),ans(n, ll(n)*(n+1)/2){}
    void built(){
        static int a,b;
        for(auto&i:c) cin>>i,--i;
        for(int i{1};i<n;++i){
            cin>>a>>b;
            --a,--b;
            e[a].emplace_back(b);
            e[b].emplace_back(a);
        }
    }
    void dfs(int i=0,int p=-1){
        ll tmp = C[c[i]];
        for(auto j:e[i]) if(j!=p){
                C[c[i]]=0;
                dfs(j,i);
                sz[i]+=sz[j];
                ans[c[i]]-=(sz[j]-C[c[i]])*(sz[j]-C[c[i]]+1)/2;
            }
        C[c[i]] = tmp+sz[i];
    }
    void solve(){
        sz.assign(n,1);
        dfs();
        for(int i{};i<n;++i) cout<<ans[i]-(n-C[i])*(n-C[i]+1)/2<<'\n';
    }
};

int main(){
    int n;
    cin>>n;
    tree t(n);
    t.built();
    t.solve();
}
### Prompt

Your task is to create 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<bits/stdc++.h>
using namespace std;
using ll = long long;

struct tree{
    size_t n,r,md;
    vector<vector<int>> e;
    vector<int> c,sz;
    vector<ll> ans,C;
    explicit tree(int n):n(n),e(n),c(n),C(n),ans(n, ll(n)*(n+1)/2){}
    void built(){
        static int a,b;
        for(auto&i:c) cin>>i,--i;
        for(int i{1};i<n;++i){
            cin>>a>>b;
            --a,--b;
            e[a].emplace_back(b);
            e[b].emplace_back(a);
        }
    }
    void dfs(int i=0,int p=-1){
        ll tmp = C[c[i]];
        for(auto j:e[i]) if(j!=p){
                C[c[i]]=0;
                dfs(j,i);
                sz[i]+=sz[j];
                ans[c[i]]-=(sz[j]-C[c[i]])*(sz[j]-C[c[i]]+1)/2;
            }
        C[c[i]] = tmp+sz[i];
    }
    void solve(){
        sz.assign(n,1);
        dfs();
        for(int i{};i<n;++i) cout<<ans[i]-(n-C[i])*(n-C[i]+1)/2<<'\n';
    }
};

int main(){
    int n;
    cin>>n;
    tree t(n);
    t.built();
    t.solve();
}
```