/*input

*/
// https://atcoder.jp/contests/abc163/tasks/abc163_f
#include <bits/stdc++.h>
#define up(i,a,b) for(int (i) = (a);(i)<=(b);++(i))
#define down(i,b,a) for(int (i) = (b);i>=(a);--i)
#define pr pair<int,int>
#define int long long
using namespace std;
const int N = 200005;
int color[N], block[N];
int ans[N];
int child[N];
int n;
int delta = 0;
vector<int> G[N];
void dfs(int u,int par){
	int preBlock = block[color[u]] + delta;
	for (auto v : G[u]){
		if (v == par) continue;
		int preVal = block[color[u]] + delta; 
		dfs(v, u);
		int newVal = block[color[u]] + delta - preVal;
		ans[color[u]] += newVal * (newVal + 1)/2;	
	}
	delta++;
	block[color[u]] = preBlock - delta;
}
signed main(){
	ios_base::sync_with_stdio(0); cin.tie(0);
	cin >> n;
	for(int i=1;i<=n;++i){
		cin >> color[i];
	}
	for(int i=1;i<n;++i){
		int u,v; cin >> u >> v;
		G[u].push_back(v);
		G[v].push_back(u);
	}
	dfs(1, 0);
	// up(i,1,n) cout << block[i] << ' ';
	for(int i=1;i<=n;++i){
		block[i] += n;
		ans[i] += 1LL * block[i] * (block[i] + 1)/2;
		ans[i] = n*(n+1)/2 - ans[i];
	}
	for(int i=1;i<=n;++i) cout << ans[i] << '\n';
}
### Prompt

Create a solution in cpp for 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
/*input

*/
// https://atcoder.jp/contests/abc163/tasks/abc163_f
#include <bits/stdc++.h>
#define up(i,a,b) for(int (i) = (a);(i)<=(b);++(i))
#define down(i,b,a) for(int (i) = (b);i>=(a);--i)
#define pr pair<int,int>
#define int long long
using namespace std;
const int N = 200005;
int color[N], block[N];
int ans[N];
int child[N];
int n;
int delta = 0;
vector<int> G[N];
void dfs(int u,int par){
	int preBlock = block[color[u]] + delta;
	for (auto v : G[u]){
		if (v == par) continue;
		int preVal = block[color[u]] + delta; 
		dfs(v, u);
		int newVal = block[color[u]] + delta - preVal;
		ans[color[u]] += newVal * (newVal + 1)/2;	
	}
	delta++;
	block[color[u]] = preBlock - delta;
}
signed main(){
	ios_base::sync_with_stdio(0); cin.tie(0);
	cin >> n;
	for(int i=1;i<=n;++i){
		cin >> color[i];
	}
	for(int i=1;i<n;++i){
		int u,v; cin >> u >> v;
		G[u].push_back(v);
		G[v].push_back(u);
	}
	dfs(1, 0);
	// up(i,1,n) cout << block[i] << ' ';
	for(int i=1;i<=n;++i){
		block[i] += n;
		ans[i] += 1LL * block[i] * (block[i] + 1)/2;
		ans[i] = n*(n+1)/2 - ans[i];
	}
	for(int i=1;i<=n;++i) cout << ans[i] << '\n';
}
```