#include<bits/stdc++.h>
#define ll long long
#define maxn 200005
#define endl '\n'
using namespace std;
int col[maxn];
vector<int> G[maxn];
ll cal(int n) {
	return (ll)n * (n + 1) / 2;
}
int sz[maxn];
ll cnt[maxn];
ll ans[maxn];
void dfs(int u,int pa) {
	sz[u] = 1;
	ll tem = cnt[col[u]];
	cnt[col[u]] = 0;
	for(int v : G[u])
		if(v != pa) {
			dfs(v,u);
			sz[u] += sz[v];
			ans[col[u]] -= cal(sz[v] - cnt[col[u]]);
			cnt[col[u]] = 0;
		}
	cnt[col[u]] = tem + sz[u];
}
int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	int n;
	cin >> n;
	for(int i=1;i<=n;i++)
		cin >> col[i];
	for(int i=1,u,v;i<n;i++) {
		cin >> u >> v;
		G[u].push_back(v);
		G[v].push_back(u);
	}
	fill(ans+1,ans+1+n,cal(n));
	dfs(1,1);
	for(int i=1;i<=n;i++)
		cout << 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>
#define ll long long
#define maxn 200005
#define endl '\n'
using namespace std;
int col[maxn];
vector<int> G[maxn];
ll cal(int n) {
	return (ll)n * (n + 1) / 2;
}
int sz[maxn];
ll cnt[maxn];
ll ans[maxn];
void dfs(int u,int pa) {
	sz[u] = 1;
	ll tem = cnt[col[u]];
	cnt[col[u]] = 0;
	for(int v : G[u])
		if(v != pa) {
			dfs(v,u);
			sz[u] += sz[v];
			ans[col[u]] -= cal(sz[v] - cnt[col[u]]);
			cnt[col[u]] = 0;
		}
	cnt[col[u]] = tem + sz[u];
}
int main() {
	ios::sync_with_stdio(0);
	cin.tie(0);
	int n;
	cin >> n;
	for(int i=1;i<=n;i++)
		cin >> col[i];
	for(int i=1,u,v;i<n;i++) {
		cin >> u >> v;
		G[u].push_back(v);
		G[v].push_back(u);
	}
	fill(ans+1,ans+1+n,cal(n));
	dfs(1,1);
	for(int i=1;i<=n;i++)
		cout << ans[i] - cal(n - cnt[i]) << endl;
	return 0;
}
```