#include<cstdio>
#include<vector>

typedef long long ll;
const int maxn = 2E+5 + 5;

int n, c[maxn];
ll siz[maxn], last[maxn], ans[maxn];
std::vector<int> to[maxn];

void DFS1(int u, int father)
{
	siz[u] = 1;
	for(int i = 0; i < to[u].size(); ++i) {
		int v = to[u][i];
		
		if(v != father) {
			DFS1(v, u);
			siz[u] += siz[v];
		}
	}
}

void DFS2(int u, int father)
{
	ans[c[u]] += (last[c[u]] - siz[u] + 1) * siz[u];
	
	ll tmp = last[c[u]] - siz[u], down = 0, T;
	
	T = last[c[u]] = siz[u] - 1;
	for(int i = 0; i < to[u].size(); ++i) {
		int v = to[u][i];
		
		if(v != father)
			DFS2(v, u), ans[c[u]] += down * (siz[v] - (T - last[c[u]])), down += siz[v] - (T - last[c[u]]), T = last[c[u]];
	}
	last[c[u]] = tmp;
}

int main() {
	scanf("%d", &n);
	for(int i = 1; i <= n; ++i) {
		last[i] = n;
		scanf("%d", &c[i]);
	}
	for(int i = 1; i < n; ++i) {
		int x, y;
		
		scanf("%d%d", &x, &y);
		to[x].push_back(y), to[y].push_back(x);
	}
	
	DFS1(1, 0), DFS2(1, 0);
	
	for(int i = 1; i <= n; ++i)
		printf("%lld\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<cstdio>
#include<vector>

typedef long long ll;
const int maxn = 2E+5 + 5;

int n, c[maxn];
ll siz[maxn], last[maxn], ans[maxn];
std::vector<int> to[maxn];

void DFS1(int u, int father)
{
	siz[u] = 1;
	for(int i = 0; i < to[u].size(); ++i) {
		int v = to[u][i];
		
		if(v != father) {
			DFS1(v, u);
			siz[u] += siz[v];
		}
	}
}

void DFS2(int u, int father)
{
	ans[c[u]] += (last[c[u]] - siz[u] + 1) * siz[u];
	
	ll tmp = last[c[u]] - siz[u], down = 0, T;
	
	T = last[c[u]] = siz[u] - 1;
	for(int i = 0; i < to[u].size(); ++i) {
		int v = to[u][i];
		
		if(v != father)
			DFS2(v, u), ans[c[u]] += down * (siz[v] - (T - last[c[u]])), down += siz[v] - (T - last[c[u]]), T = last[c[u]];
	}
	last[c[u]] = tmp;
}

int main() {
	scanf("%d", &n);
	for(int i = 1; i <= n; ++i) {
		last[i] = n;
		scanf("%d", &c[i]);
	}
	for(int i = 1; i < n; ++i) {
		int x, y;
		
		scanf("%d%d", &x, &y);
		to[x].push_back(y), to[y].push_back(x);
	}
	
	DFS1(1, 0), DFS2(1, 0);
	
	for(int i = 1; i <= n; ++i)
		printf("%lld\n", ans[i]);
}
```