//
//  main.cpp
//  F
//
//  Created by 曾憲揚 on 2020/8/19.
//  Copyright © 2020 曾憲揚. All rights reserved.
//

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

typedef long long ll;
const int maxn = 2e5+5;

int c[maxn];
vector<int> G[maxn];
ll sol[maxn];
int n;
int cnt_son[maxn];
int num[maxn];

ll cal(int x){
    return 1LL*x*(x+1)/2;
}

void DFS(int p, int cur){
    cnt_son[cur] = 1;
    int col = c[cur];
    int tmp = num[col];
    for(int i:G[cur]){
        if(i!=p){
            int now = num[col];
            DFS(cur, i);
            now = num[col]-now;
            sol[col] -= cal(cnt_son[i]-now);
            cnt_son[cur] += cnt_son[i];
        }
    }
    num[col] = tmp + cnt_son[cur];
}

int main(int argc, const char * argv[]) {
    ios_base::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);
    
    cin>>n;
    for(int i=1; i<=n; i++) cin>>c[i];
    for(int i=1; i<n; i++){
        int a, b; cin>>a>>b;
        G[a].push_back(b); G[b].push_back(a);
    }
    fill(sol+1, sol+1+n, cal(n));
    
    DFS(0, 1);
    for(int i=1; i<=n; i++){
        sol[i] -= cal(n-num[i]);
        cout<<sol[i]<<" ";
    }
    cout<<endl;
    return 0;
}

### 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
//
//  main.cpp
//  F
//
//  Created by 曾憲揚 on 2020/8/19.
//  Copyright © 2020 曾憲揚. All rights reserved.
//

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

typedef long long ll;
const int maxn = 2e5+5;

int c[maxn];
vector<int> G[maxn];
ll sol[maxn];
int n;
int cnt_son[maxn];
int num[maxn];

ll cal(int x){
    return 1LL*x*(x+1)/2;
}

void DFS(int p, int cur){
    cnt_son[cur] = 1;
    int col = c[cur];
    int tmp = num[col];
    for(int i:G[cur]){
        if(i!=p){
            int now = num[col];
            DFS(cur, i);
            now = num[col]-now;
            sol[col] -= cal(cnt_son[i]-now);
            cnt_son[cur] += cnt_son[i];
        }
    }
    num[col] = tmp + cnt_son[cur];
}

int main(int argc, const char * argv[]) {
    ios_base::sync_with_stdio(0);
    cin.tie(0); cout.tie(0);
    
    cin>>n;
    for(int i=1; i<=n; i++) cin>>c[i];
    for(int i=1; i<n; i++){
        int a, b; cin>>a>>b;
        G[a].push_back(b); G[b].push_back(a);
    }
    fill(sol+1, sol+1+n, cal(n));
    
    DFS(0, 1);
    for(int i=1; i<=n; i++){
        sol[i] -= cal(n-num[i]);
        cout<<sol[i]<<" ";
    }
    cout<<endl;
    return 0;
}

```