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 LL long long
const int N = 2e5 + 10;
int n, c[N], cnt[N], siz[N];
	inline LL count(int n) {
		// C(n,2) + n
		return 1ll*(n*(n+1))/2;
	}
LL ans[N];
vector<int>G[N];
	void dfs(int v, int fa) {
		int tmp = cnt[c[v]];
		siz[v] = 1;
		for(int u:G[v]) {
			if(u==fa) continue;
			cnt[c[v]] = 0;
			dfs(u, v);
			siz[v] += siz[u];
			ans[c[v]]+=count(siz[u]-cnt[c[v]]);
		}
		cnt[c[v]] = tmp + siz[v];
	}
int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; i++)scanf("%d", &c[i]);
	for (int i = 1; i < n; i++) {
		int u, v;
		scanf("%d%d", &u, &v);
		G[u].push_back(v);
		G[v].push_back(u);
	}
	dfs(1, 0);
	for (int i = 1; i <= n; i++) {
		printf("%lld\n", count(n) - ans[i] - count(n - cnt[i]));
	}
	return 0;
}