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 < int(n); i++)
#define rrep(i, n) for (int i = int(n) - 1; i >= 0; i--)
#define reps(i, n) for (int i = 1; i <= int(n); i++)
#define rreps(i, n) for (int i = int(n); i >= 1; i--)
#define repc(i, n) for (int i = 0; i <= int(n); i++)
#define rrepc(i, n) for (int i = int(n); i >= 0; i--)
#define repi(i, a, b) for (int i = int(a); i < int(b); i++)
#define repic(i, a, b) for (int i = int(a); i <= int(b); i++)
#define all(a) (a).begin(), (a).end()
#define bit32(x) (1 << (x))
#define bit64(x) (1ll << (x))
#define sz(v) ((int) v.size())

using namespace std;

using i64 = long long;
using f80 = long double;
using vi32 = vector<int>;
using vi64 = vector<i64>;
using vf80 = vector<f80>;
using vstr = vector<string>;

inline void yes() { cout << "Yes" << endl; exit(0); }
inline void no() { cout << "No" << endl; exit(0); }
inline i64 gcd(i64 a, i64 b) { if (min(a, b) == 0) return max(a, b); if (a % b == 0) return b; return gcd(b, a % b); }
inline i64 lcm(i64 a, i64 b) { if (min(a, b) == 0) return max(a, b); return a / gcd(a, b) * b; }
template <typename T> class pqasc : public priority_queue<T, vector<T>, greater<T>> {};
template <typename T> class pqdesc : public priority_queue<T, vector<T>, less<T>> {};
template <typename T> inline void amax(T &x, T y) { x = max(x, y); }
template <typename T> inline void amin(T &x, T y) { x = min(x, y); }
template <typename T> inline T exp(T x, i64 n, T e = 1) { T r = e; while (n > 0) { if (n & 1) r *= x; x *= x; n >>= 1; } return r; }
template <typename T> istream& operator>>(istream &is, vector<T> &v) { for (auto &x : v) is >> x; return is; }
template <typename T> ostream& operator<<(ostream &os, vector<T> &v) { rep(i, v.size()) { if (i) os << ' '; os << v[i]; } return os; }
void solve(); int main() { ios::sync_with_stdio(0); cin.tie(0); cout << fixed << setprecision(16); solve(); return 0; }

template <typename T, typename S>
struct LazySegmentTree {
  using F = function<T(T, T)>;
  using G = function<T(T, S, int)>;
  using H = function<S(S, S)>;
  vector<T> v;
  vector<S> z;
  F f;
  G g;
  H h;
  T e;
  S d;
  int n;
  LazySegmentTree(int size, F f, G g, H h, T e, S d) : f(f), g(g), h(h), e(e), d(d) {
    n = 1;
    while (n < size) n <<= 1;
    v.resize(n * 2, e);
    z.resize(n * 2, d);
  }
  void set(int k, T x) {
    v[k + n] = x;
  }
  void build() {
    rrep(i, n) v[i] = f(v[i * 2 + 0], v[i * 2 + 1]);
  }
  void propagate(int k, int l, int r) {
    v[k] = g(v[k], z[k], r - l);
    if (k < n) {
      z[k * 2 + 0] = h(z[k * 2 + 0], z[k]);
      z[k * 2 + 1] = h(z[k * 2 + 1], z[k]);
    }
    z[k] = d;
  }
  T update(int a, int b, S x, int k = 1, int l = 0, int r = -1) {
    if (r == -1) r = n;
    propagate(k, l, r);
    if (b <= l || r <= a) return v[k];
    if (a <= l && r <= b) { z[k] = x; propagate(k, l, r); return v[k]; }
    return v[k] = f(
      update(a, b, x, k * 2 + 0, l, (l + r) / 2),
      update(a, b, x, k * 2 + 1, (l + r) / 2, r)
    );
  }
  T query(int a, int b, int k = 1, int l = 0, int r = -1) {
    if (r == -1) r = n;
    propagate(k, l, r);
    if (b <= l || r <= a) return e;
    if (a <= l && r <= b) return v[k];
    return f(
      query(a, b, k * 2 + 0, l, (l + r) / 2),
      query(a, b, k * 2 + 1, (l + r) / 2, r)
    );
  }
};

void solve() {
  int n;
  cin >> n;
  vi32 c(n);
  cin >> c;
  rep(i, n) {
    c[i]--;
  }
  vector<vi32> g(n);
  rep(i, n - 1) {
    int a, b;
    cin >> a >> b;
    a--, b--;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  LazySegmentTree<int, int> lst(n,
    [](int a, int b) { return a + b; },
    [](int a, int b, int k) { return a + b * k; },
    [](int a, int b) { return a + b; },
    0, 0);
  vector<vi32> cnts(n);
  auto dfs = [&](auto dfs, int u, int p) -> void {
    int co = c[u];
    int old = lst.query(co, co + 1);
    lst.update(0, n, 1);
    for (auto v : g[u]) if (v != p) {
      int old = lst.query(co, co + 1);
      lst.update(co, co + 1, -old);
      dfs(dfs, v, u);
      int now = lst.query(co, co + 1);
      if (now > 0) {
        cnts[co].push_back(now);
      }
    }
    int now = lst.query(co, co + 1);
    lst.update(co, co + 1, -now);
    lst.update(co, co + 1, old);
  };
  dfs(dfs, 0, -1);
  rep(i, n) {
    int c = lst.query(i, i + 1);
    if (c > 0) {
      cnts[i].push_back(c);
    }
  }
  rep(i, n) {
    i64 ans = 0;
    for (auto c : cnts[i]) {
      ans += (i64) c * (c - 1) / 2 + c;
    }
    ans = (i64) n * (n - 1) / 2 + n - ans;
    cout << ans << '\n';
  }
}