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>
#define rep(i,n) for (int i = 0; i < (n); ++i)
using namespace std;
using ll = long long;
using P = pair<int,int>;
 
template<typename T>
struct BIT {
  int n;
  vector<T> d;
  BIT(int n=0):n(n),d(n+1) {}
  void add(int i, T x=1) {
    for (i++; i <= n; i += i&-i) {
      d[i] += x;
    }
  }
  T sum(int i) {
    T x = 0;
    for (i++; i; i -= i&-i) {
      x += d[i];
    }
    return x;
  }
  T sum(int l, int r) {
    return sum(r-1) - sum(l-1);
  }
};
 
const int MX = 200005;
int in[MX], out[MX], k;
vector<int> to[MX];
void dfs(int v, int p=-1) {
  in[v] = k++;
  for (int u : to[v]) {
    if (u == p) continue;
    dfs(u,v);
  }
  out[v] = k;
}
 
vector<int> vs[MX];
 
ll c2(ll n) { return n*(n+1)/2;}
 
int main() {
  int n;
  cin >> n;
  vector<int> c(n);
  rep(i,n) cin >> c[i];
  rep(i,n-1) {
    int a, b;
    cin >> a >> b;
    --a; --b;
    to[a].push_back(b);
    to[b].push_back(a);
  }
  dfs(0);
  rep(i,n) vs[c[i]-1].push_back(i);
  ll tot = c2(n);
  BIT<int> d(n);
  rep(i,n) d.add(i,1);
  rep(i,n) {
    ll ans = tot;
    vector<int>& nvs = vs[i];
    sort(nvs.begin(), nvs.end(), [&](int a, int b) {
      return in[a] > in[b];
    });
    vector<P> history;
    for (int v : nvs) {
      int cnt = 1;
      for (int u : to[v]) {
        if (in[u] < in[v]) continue;
        int num = d.sum(in[u], out[u]);
        ans -= c2(num);
        cnt += num;
      }
      d.add(in[v], -cnt);
      history.emplace_back(in[v], cnt);
    }
    ans -= c2(d.sum(0,n));
    for (auto p : history) {
      d.add(p.first, p.second);
    }
    cout << ans << endl;
  }
  return 0;
}