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: Accepted)

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

typedef long long ll;
typedef long double ld;

int col[200005];
vector <int> graf[200005];
int posl[200005];
int last[200005];
vector <int> novi[200005];
ll res[200005];
ll sz[200005];
bool ima[200005];
ll sve[200005];
int par[200005];
int sad[200005];

void dfs_init(int v, int parent){
    par[v] = parent;
    last[v] = posl[col[v]];
    sz[v] = 1;
    posl[col[v]] = v;
    for(auto c : graf[v]){
        if(c == parent) continue;
        sad[v] = c;
        dfs_init(c, v);
        posl[col[v]] = v;
        sz[v] += sz[c];
    }
    posl[col[v]] = last[v];
    novi[sad[last[v]]].push_back(v);
}

int main(){
    ios_base::sync_with_stdio(false), cin.tie(0);
    cout.precision(10);
    cout << fixed;

    int n;
    cin >> n;
    for(int i=1; i<=n; i++){
        cin >> col[i];
        ima[col[i]] = 1;
        sve[i] = n;
    }
    for(int i=1; i<n; i++){
        int a, b;
        cin >> a >> b;
        graf[a].push_back(b);
        graf[b].push_back(a);
    }
    dfs_init(1, 0);
    for(int i=1; i<=n; i++){
        if(last[i] == 0) sve[col[i]] -= sz[i];
    }
    for(int i=1; i<=n; i++){
        ll tr = sz[i];
        for(auto c : novi[i]){
            tr -= sz[c];
        }
        res[col[par[i]]] += tr*(tr+1)/2;
    }
    for(int i=1; i<=n; i++){
        res[i] += 1LL*sve[i]*(sve[i]+1)/2;
        cout << 1LL*n*(n+1)/2 - res[i] << "\n";
    }
    return 0;
}