Problem:
Score : 600 points

Problem Statement

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.

Sample Input 1

3
1 2 1
1 2
2 3

Sample Output 1

5
4
0

Let P_{i,j} denote the simple path connecting Vertex i and j.

There are 5 simple paths that visit a vertex painted in the color 1 one or more times:

P_{1,1}\,,\,
P_{1,2}\,,\,
P_{1,3}\,,\,
P_{2,3}\,,\,
P_{3,3}

There are 4 simple paths that visit a vertex painted in the color 2 one or more times:

P_{1,2}\,,\,
P_{1,3}\,,\,
P_{2,2}\,,\,
P_{2,3}

There are no simple paths that visit a vertex painted in the color 3 one or more times.

Sample Input 2

1
1

Sample Output 2

1

Sample Input 3

2
1 2
1 2

Sample Output 3

2
2

Sample Input 4

5
1 2 3 4 5
1 2
2 3
3 4
3 5

Sample Output 4

5
8
10
5
5

Sample Input 5

8
2 7 2 5 4 1 7 5
3 1
1 2
2 7
4 5
5 6
6 8
7 8

Sample Output 5

18
15
0
14
23
0
23
0

Solution in C++ (GCC 9.2.1) (judge verdict: Wrong Answer)

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

#define read(num) scanf("%d", &num)
#define readll(num) scanf("%lld", &num)
#define read2(num1, num2) scanf("%d%d", &num1, &num2)
#define read3(num1, num2, num3) scanf("%d%d%d", &num1, &num2, &num3)
#define reads(str) scanf("%s", str)
#define readc(ch) scanf("%c", &ch)

#define pb push_back
#define mp make_pair

#define pii pair<int, int>
#define vi vector<int>
#define ll long long
ll color[200005],child[200005],ans[200005];
vector<ll>adj[200005];

ll n;
void dfs(int in,int vis[]){
    vis[in]=1;
    child[in]=1;
    vector<ll>kk;
    ans[color[in]]+=1;
    for(int i=0;i<adj[in].size();i++){
        int v=adj[in][i];
        if(vis[v]==0){
            dfs(v,vis);
            kk.pb(child[v]);
            child[in]+=child[v];
        }
    }
    ans[color[in]]+=(n-child[in])*child[in];
    for(int i=0;i<kk.size();i++){
        for(int j=i+1;j<kk.size();j++){
            ans[color[in]]+=kk[i]*kk[j];
        }
        ans[color[in]]+=kk[i];
    }
}
int main()
{
   
    cin>>n;
    ll c[n+1]={0};
    for(int i=1;i<=n;i++){
        cin>>color[i];
        c[color[i]]++;
    }
    ll u,v;
    for(int i=1;i<n;i++){
        cin>>u>>v;
        adj[u].pb(v);
        adj[v].pb(u);
    }
    int vis[n+1]={0};
    dfs(1,vis);
    for(int i=1;i<=n;i++){
        ans[i]-=((c[i]-1)*((c[i]-1)*c[i])/2);
        cout<<ans[i]<<endl;
    }
}