#include<bits/stdc++.h>
using namespace std;
const int N = 2e5+5;
using ll = long long;

vector<ll> g[N];
vector<ll> dist(N),col(N);
vector<ll> res(N),c(N);
ll f(ll n){
    return 1LL*n*(n+1)/2;
}

void dfs(ll x, ll p){
    ll t = c[col[x]];
    for(auto i : g[x]){
        if(i==p) continue;
        c[col[x]] = 0;
        dfs(i,x);
        dist[x]+=dist[i];
        res[col[x]] += f(dist[i]-c[col[x]]);
    }
    c[col[x]]=t+dist[x];
}


int main(){
    int n;
    cin>>n;
    
    for(int i = 1; i <= n; i++){
        dist[i] = 1;
        cin>>col[i];
    } 

    for(int i = 0; i < n-1; i++){
        int a,b;
        cin>>a>>b;
        g[a].push_back(b);
        g[b].push_back(a);
    }

    dfs(1,-1);

    for(int i = 1; i <= n; i++){
        cout<<f(n) - f(n-c[i]) - res[i]<<"\n";
    }
}

### Prompt

Please formulate 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;
const int N = 2e5+5;
using ll = long long;

vector<ll> g[N];
vector<ll> dist(N),col(N);
vector<ll> res(N),c(N);
ll f(ll n){
    return 1LL*n*(n+1)/2;
}

void dfs(ll x, ll p){
    ll t = c[col[x]];
    for(auto i : g[x]){
        if(i==p) continue;
        c[col[x]] = 0;
        dfs(i,x);
        dist[x]+=dist[i];
        res[col[x]] += f(dist[i]-c[col[x]]);
    }
    c[col[x]]=t+dist[x];
}


int main(){
    int n;
    cin>>n;
    
    for(int i = 1; i <= n; i++){
        dist[i] = 1;
        cin>>col[i];
    } 

    for(int i = 0; i < n-1; i++){
        int a,b;
        cin>>a>>b;
        g[a].push_back(b);
        g[b].push_back(a);
    }

    dfs(1,-1);

    for(int i = 1; i <= n; i++){
        cout<<f(n) - f(n-c[i]) - res[i]<<"\n";
    }
}

```