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)

//#pragma GCC optimize("Ofast")
//#pragma GCC target("avx,avx2,fma")

#include <bits/stdc++.h>

using namespace std;

#define int long long
#define sz(a) (int) (a).size()
#define all(a) (a).begin(), (a).end()
#define x first
#define y second

mt19937 rnd(chrono::steady_clock::now().time_since_epoch().count());

const int maxn = 2e5 + 10;

int c[maxn], sz[maxn], ans[maxn], add[maxn], cnt[maxn];
vector<int> g[maxn];
unordered_map<int, int> mp[maxn];
unordered_map<int, int> mp2;

int n;

void dfs(int v, int p = -1) {
    sz[v] = 1;
    int cnt2 = 1;
    for (auto &u : g[v]) {
        if (u == p) continue;
        dfs(u, v);
        sz[v] += sz[u];
        if (sz(mp[u]) > sz(mp[v])) {
            swap(mp[u], mp[v]);
        }
        cnt2 += sz[u];
        for (auto &el : mp[u]) {
            if (p == -1 && el.x == c[v]) {
                cnt[c[v]] += (cnt2 - mp[v][c[v]]) * el.y;
            }
            mp[v][el.x] += el.y;
        }
    }
    ans[v] += sz[v] - mp[v][c[v]];
    add[v] = sz[v] - mp[v][c[v]];
    mp[v][c[v]] = sz[v];
}

void dfs2(int v, int p = -1) {
    if (p == -1) mp2 = mp[v];
//    int was = mp2[c[v]];
    mp2[c[v]] -= add[v];
    for (auto u : g[v]) {
        if (u == p) continue;
        cnt[c[u]] += (n - mp2[c[u]]) * ans[u];
        dfs2(u, v);
    }
    mp2[c[v]] += add[v];
}

signed main() {
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);

    cin >> n;
    for (int i = 1; i <= n; i++) {
        cin >> c[i];
    }
    for (int i = 0; i < n - 1; i++) {
        int u, v;
        cin >> u >> v;
        g[u].push_back(v);
        g[v].push_back(u);
    }
    dfs(1);
    dfs2(1);
    for (int i = 1; i <= n; i++) {
        cout << cnt[i] << "\n";
    }
    cout << "\n";

    return 0;
}