#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, n) for (ll i = 0; i < (ll)(n); i++)
ll dfs(ll x, vector<vector<ll>> &g, vector<bool> &see,vector<ll> &cl,vector<ll> &ans,vector<ll> &c,vector<stack<ll>> &cn,vector<ll> &par)
{
  ll chi=0;
  //点xに対する処理
  see[x] = 1;
  ll p=-1;
  if(!cn[c[x]].empty())p=cn[c[x]].top();
  cn[c[x]].push(x);
  for (auto r : g[x])
  {
    if (see[r])continue;
    //進む辺に対する処理（これは木構造でオイラーツアーするために使う）
    cl[x]=0;
    ll chit=dfs(r, g, see,cl,ans,c,cn,par)+1;
    ans[c[x]]-=(chit+cl[x])*(chit+cl[x]+1)/2;
    //戻る辺に対する処理
    chi+=chit;
  }
  //点xの親に戻る時の処理
  if(p==-1){
    par[c[x]]-=chi+1;
  }
  else{
    cl[p]-=chi+1;
  }
    cn[c[x]].pop();
  return chi;
}

int main()
{
ll n;
cin >> n ;
vector<ll> ans(n,n*(n+1)/2);
vector<ll> cl(n,0);
vector<ll> par(n,n);
vector<stack<ll>> cn(n);
vector<bool> see(n,0);
vector<ll> c(n,0);
rep(i, n){
  ll cc;
  cin >> cc;
  c[i]=cc-1;
}
vector<vector<ll>> g(n);
rep(i, n-1)
{
  ll a, b;
  cin >> a >> b;
  g[a - 1].push_back(b - 1);
  g[b - 1].push_back(a - 1); 
}
dfs(0,g,see,cl,ans,c,cn,par);
rep(i,n)cout << ans[i]-par[i]*(par[i]+1)/2 << endl;
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 <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define rep(i, n) for (ll i = 0; i < (ll)(n); i++)
ll dfs(ll x, vector<vector<ll>> &g, vector<bool> &see,vector<ll> &cl,vector<ll> &ans,vector<ll> &c,vector<stack<ll>> &cn,vector<ll> &par)
{
  ll chi=0;
  //点xに対する処理
  see[x] = 1;
  ll p=-1;
  if(!cn[c[x]].empty())p=cn[c[x]].top();
  cn[c[x]].push(x);
  for (auto r : g[x])
  {
    if (see[r])continue;
    //進む辺に対する処理（これは木構造でオイラーツアーするために使う）
    cl[x]=0;
    ll chit=dfs(r, g, see,cl,ans,c,cn,par)+1;
    ans[c[x]]-=(chit+cl[x])*(chit+cl[x]+1)/2;
    //戻る辺に対する処理
    chi+=chit;
  }
  //点xの親に戻る時の処理
  if(p==-1){
    par[c[x]]-=chi+1;
  }
  else{
    cl[p]-=chi+1;
  }
    cn[c[x]].pop();
  return chi;
}

int main()
{
ll n;
cin >> n ;
vector<ll> ans(n,n*(n+1)/2);
vector<ll> cl(n,0);
vector<ll> par(n,n);
vector<stack<ll>> cn(n);
vector<bool> see(n,0);
vector<ll> c(n,0);
rep(i, n){
  ll cc;
  cin >> cc;
  c[i]=cc-1;
}
vector<vector<ll>> g(n);
rep(i, n-1)
{
  ll a, b;
  cin >> a >> b;
  g[a - 1].push_back(b - 1);
  g[b - 1].push_back(a - 1); 
}
dfs(0,g,see,cl,ans,c,cn,par);
rep(i,n)cout << ans[i]-par[i]*(par[i]+1)/2 << endl;
return 0;
}
```