#include <bits/stdc++.h>

using namespace std;

int n, a[200005], siz[200005], cnt[400005];
long long ans[200005];
vector <int> edge[200005];
stack <int> stk[200005];

long long f(int x) {
	return 1LL * x * (x + 1) / 2;
}

void calc(int id, int p) {
	siz[id] = 1;
	for (int i = 0; i < edge[id].size(); i++) {
		int to = edge[id][i];
		if (to == p) continue;
		calc(to, id);
		siz[id] += siz[to];
	}
}

void dfs(int id, int p) {
	int before = stk[a[id]].top();
	cnt[before] -= siz[id];
	
	stk[a[id]].push(id);
	
	for (int i = 0; i < edge[id].size(); i++) {
		int to = edge[id][i];
		if (to == p) continue;
		cnt[id] = siz[to];
		dfs(to, id);
		ans[a[id]] += f(cnt[id]);
	}
	
	stk[a[id]].pop();
}

int main() {
	scanf("%d", &n);
	
	for (int i = 0; i < n; i++) scanf("%d", &a[i]);
	
	for (int i = 0; i < n - 1; i++) {
		int x, y;
		scanf("%d%d", &x, &y);
		x--;
		y--;
		edge[x].push_back(y);
		edge[y].push_back(x);
	}
	
	for (int i = 1; i <= n; i++) {
		cnt[n - 1 + i] = n;
		stk[i].push(n - 1 + i);
	}
	
	calc(0, -1);
	dfs(0, -1);
	
	for (int i = 1; i <= n; i++) {
		ans[i] += f(cnt[n - 1 + i]);
		printf("%lld\n", f(n) - ans[i]);
	}
}

### Prompt

Develop a solution in Cpp to the problem described below:

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;

int n, a[200005], siz[200005], cnt[400005];
long long ans[200005];
vector <int> edge[200005];
stack <int> stk[200005];

long long f(int x) {
	return 1LL * x * (x + 1) / 2;
}

void calc(int id, int p) {
	siz[id] = 1;
	for (int i = 0; i < edge[id].size(); i++) {
		int to = edge[id][i];
		if (to == p) continue;
		calc(to, id);
		siz[id] += siz[to];
	}
}

void dfs(int id, int p) {
	int before = stk[a[id]].top();
	cnt[before] -= siz[id];
	
	stk[a[id]].push(id);
	
	for (int i = 0; i < edge[id].size(); i++) {
		int to = edge[id][i];
		if (to == p) continue;
		cnt[id] = siz[to];
		dfs(to, id);
		ans[a[id]] += f(cnt[id]);
	}
	
	stk[a[id]].pop();
}

int main() {
	scanf("%d", &n);
	
	for (int i = 0; i < n; i++) scanf("%d", &a[i]);
	
	for (int i = 0; i < n - 1; i++) {
		int x, y;
		scanf("%d%d", &x, &y);
		x--;
		y--;
		edge[x].push_back(y);
		edge[y].push_back(x);
	}
	
	for (int i = 1; i <= n; i++) {
		cnt[n - 1 + i] = n;
		stk[i].push(n - 1 + i);
	}
	
	calc(0, -1);
	dfs(0, -1);
	
	for (int i = 1; i <= n; i++) {
		ans[i] += f(cnt[n - 1 + i]);
		printf("%lld\n", f(n) - ans[i]);
	}
}

```