#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 cal(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]]+=cal(siz[u]-cnt[c[v]]);
		}
		cnt[c[v]] = tmp + siz[v];
	}
	
	#define ADD(a, b) 	\
		do {			\
			G[a].push_back(b);	\
			G[b].push_back(a);	\
		}while(0);		
		
	inline void add(int a, int b) {
		G[a].push_back(b);
		G[b].push_back(a);
	}
	
int main() {
	//scanf("%d", &n);
	std::ios::sync_with_stdio(false);
	cin>>n;
	for (int i = 1; i <= n; i++)
		//scanf("%d", &c[i]);
		cin>>c[i];
	for (int i = 1; i < n; i++) {
		int u, v;
		//scanf("%d%d", &u, &v);
		cin>>u>>v;
		G[u].push_back(v);
		G[v].push_back(u);
		//ADD(u, v); ADD(v, u);
	}
	dfs(1, 0);
	for (int i = 1; i <= n; i++) {
		cout<< cal(n) - ans[i] - cal(n - cnt[i])<<endl;
	}
	return 0;
}

### Prompt

Your task is to create a cpp solution to the following problem:

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.

Examples

Input

3
1 2 1
1 2
2 3


Output

5
4
0


Input

1
1


Output

1


Input

2
1 2
1 2


Output

2
2


Input

5
1 2 3 4 5
1 2
2 3
3 4
3 5


Output

5
8
10
5
5


Input

8
2 7 2 5 4 1 7 5
3 1
1 2
2 7
4 5
5 6
6 8
7 8


Output

18
15
0
14
23
0
23
0

### Response

```cpp
#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 cal(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]]+=cal(siz[u]-cnt[c[v]]);
		}
		cnt[c[v]] = tmp + siz[v];
	}
	
	#define ADD(a, b) 	\
		do {			\
			G[a].push_back(b);	\
			G[b].push_back(a);	\
		}while(0);		
		
	inline void add(int a, int b) {
		G[a].push_back(b);
		G[b].push_back(a);
	}
	
int main() {
	//scanf("%d", &n);
	std::ios::sync_with_stdio(false);
	cin>>n;
	for (int i = 1; i <= n; i++)
		//scanf("%d", &c[i]);
		cin>>c[i];
	for (int i = 1; i < n; i++) {
		int u, v;
		//scanf("%d%d", &u, &v);
		cin>>u>>v;
		G[u].push_back(v);
		G[v].push_back(u);
		//ADD(u, v); ADD(v, u);
	}
	dfs(1, 0);
	for (int i = 1; i <= n; i++) {
		cout<< cal(n) - ans[i] - cal(n - cnt[i])<<endl;
	}
	return 0;
}

```