#include <bits/stdc++.h>
using namespace std;
#define all(V) V.begin(),V.end()
#define ll long long
#define MOD 1000000007

vector<int> G[200010];
stack<ll> S[200010];

ll alc = 0, N;
ll idc[200010], C[200010], ans[200010], ap[200010];

void euler(int c, int p) {
	for (int i : G[c]) {
		if (i == p)continue;
		S[C[c]].push(alc + idc[C[c]]);
		idc[C[c]] = -alc;
		alc++;
		idc[C[i]]--;
		euler(i, c);
	}
	ll g;
	if (p < 0)return;
	if (p == 0) {
		for (int i = 1;i <= N;i++) {
			g = idc[i] + alc;
			ans[i] += g * (g - 1) / 2;
		}
		return;
	}
	g = idc[C[p]] + alc;
	ans[C[p]] += g * (g - 1) / 2;
	idc[C[p]] = S[C[p]].top() - alc;
	S[C[p]].pop();
}


int main() {
	for (int i = 0;i < 200010;i++)idc[i] = C[i] = ans[i] = ap[i] = 0;
	int a, b;
	cin >> N;
	for (int i = 1;i <= N;i++) {
		cin >> C[i];
		ap[C[i]]++;
	}
	for (int i = 1;i < N;i++) {
		cin >> a >> b;
		G[a].push_back(b);
		G[b].push_back(a);
	}
	G[0] = { 1 };
	S[0].push(0);
	euler(0, -1);
	for (int i = 1;i <= N;i++)
		if (ap[i] == 0)cout << 0 << endl;
		else cout << N * (N - 1) / 2 - ans[i] + ap[i] << endl;
}
### 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 all(V) V.begin(),V.end()
#define ll long long
#define MOD 1000000007

vector<int> G[200010];
stack<ll> S[200010];

ll alc = 0, N;
ll idc[200010], C[200010], ans[200010], ap[200010];

void euler(int c, int p) {
	for (int i : G[c]) {
		if (i == p)continue;
		S[C[c]].push(alc + idc[C[c]]);
		idc[C[c]] = -alc;
		alc++;
		idc[C[i]]--;
		euler(i, c);
	}
	ll g;
	if (p < 0)return;
	if (p == 0) {
		for (int i = 1;i <= N;i++) {
			g = idc[i] + alc;
			ans[i] += g * (g - 1) / 2;
		}
		return;
	}
	g = idc[C[p]] + alc;
	ans[C[p]] += g * (g - 1) / 2;
	idc[C[p]] = S[C[p]].top() - alc;
	S[C[p]].pop();
}


int main() {
	for (int i = 0;i < 200010;i++)idc[i] = C[i] = ans[i] = ap[i] = 0;
	int a, b;
	cin >> N;
	for (int i = 1;i <= N;i++) {
		cin >> C[i];
		ap[C[i]]++;
	}
	for (int i = 1;i < N;i++) {
		cin >> a >> b;
		G[a].push_back(b);
		G[b].push_back(a);
	}
	G[0] = { 1 };
	S[0].push(0);
	euler(0, -1);
	for (int i = 1;i <= N;i++)
		if (ap[i] == 0)cout << 0 << endl;
		else cout << N * (N - 1) / 2 - ans[i] + ap[i] << endl;
}
```