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<iostream>
	#include<vector>
	using namespace std;
	#define LL	long long
	const int MAXN = 2e5+10;

	int N;
	int c[MAXN];
	int tot = 0;
	vector<int> G[MAXN];
	//int head[MAXN], Next[MAXN], to[MAXN];

	// sub[i]: 顶点i下子树顶点的个数
	int sub[MAXN];
	// cnt[i]: 颜色为i的顶点下子树顶点的个数
	int cnt[MAXN]; 
	// ans[i]: 颜色为i的顶点下不经过颜色i的顶点数
	LL ans[MAXN];
	inline LL count(int n) {
		// C(n,2) + n
		return 1LL*(n*(n+1))/2;
	}
	/*
	void dfs(int v, int fa) {
		int tmp = cnt[c[v]];
		sub[v] = 1;
		for(int i=head[v]; i; i=Next[i]) {
			int u = to[i];
			if(u == fa) continue;
			cnt[c[v]] = 0;	// 置0重新计数
			dfs(u, v);
			sub[v] += sub[u];
			ans[c[v]] += count(sub[u] - cnt[c[v]]);
		}
		cnt[c[v]] = tmp + sub[v];
	}
	void add(int a, int b) {
		++tot;
		Next[tot] = head[a];
		to[tot] = b;
		head[a] = tot;
	}
	*/
	void dfs(int v, int fa) {
		int tmp = cnt[c[v]];
		sub[v] = 1;
		for(int u:G[v]) {
			if(u==fa) continue;
			cnt[c[v]] = 0;
			dfs(u, v);
			sub[v] += sub[u];
			ans[c[v]]+=count(sub[u]-cnt[c[v]]);
		}
		cnt[c[v]] = tmp + sub[v];
	}
	
	void add(int a, int b) {
		G[a].push_back(b);
		G[b].push_back(a);
	}
	int main() {
		cin>>N;
		for(int i=1; i<=N; i++) {
			cin >> c[i];
		}
		int a, b;
		for(int i=1; i<N; i++) {
			cin>>a>>b;
			add(a, b);
			add(b, a);
		}
		//dfs1(1, 0);
		dfs(1, 0);
		for(int i=1; i<=N; i++) {
			cout << count(N) - ans[i] - count(N-cnt[i]) << endl;
		}
		return 0;
	}