#include <stdio.h>
#include <algorithm>
#include <vector>
using namespace std;

int N, K, S[200200], E[200200], Z[200200], B[200200];
vector<int> V[200200], I[200200];

void go(int x, int l)
{
	S[x] = ++K;
	for (auto &y : V[x]) if (y != l){
		go(y, x);
	}
	E[x] = K;
	Z[x] = E[x] - S[x] + 1;
}

void push(int x, int p)
{
	while (x <= N){
		B[x] += p;
		x += x & (-x);
	}
}

int pop(int x)
{
	int r = 0;
	while (x){
		r += B[x];
		x -= x & (-x);
	}
	return r;
}

int range(int x)
{
	return pop(E[x]) - pop(S[x] - 1);
}

int main()
{
	scanf ("%d", &N);
	for (int i = 1, c; i <= N; i++){
		scanf ("%d", &c);
		I[c].push_back(i);
	}
	for (int i = 1; i < N; i++){
		int x, y; scanf ("%d %d", &x, &y);
		V[x].push_back(y);
		V[y].push_back(x);
	}
	go(1, 0);

	for (int c = 1; c <= N; c++){
		long long ans = 1ll * N * (N + 1) / 2;

		sort(I[c].begin(), I[c].end(), [&](int x, int y){return S[x] > S[y]; });
		vector<pair<int, int> > q;
		for (int x : I[c]){
			int e = 1;
			for (int y : V[x]) if (S[x] < S[y]){
				int z = Z[y] - range(y);
				ans -= 1ll * z * (z + 1) / 2;
				e += z;
			}
			push(S[x], e);
			q.push_back({ S[x], e });
		}

		int z = N - range(1);
		ans -= 1ll * z * (z + 1) / 2;
		for (auto &p : q) push(p.first, -p.second);
		printf ("%lld\n", ans);
	}

	return 0;
}
### Prompt

Your challenge is to write 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 <stdio.h>
#include <algorithm>
#include <vector>
using namespace std;

int N, K, S[200200], E[200200], Z[200200], B[200200];
vector<int> V[200200], I[200200];

void go(int x, int l)
{
	S[x] = ++K;
	for (auto &y : V[x]) if (y != l){
		go(y, x);
	}
	E[x] = K;
	Z[x] = E[x] - S[x] + 1;
}

void push(int x, int p)
{
	while (x <= N){
		B[x] += p;
		x += x & (-x);
	}
}

int pop(int x)
{
	int r = 0;
	while (x){
		r += B[x];
		x -= x & (-x);
	}
	return r;
}

int range(int x)
{
	return pop(E[x]) - pop(S[x] - 1);
}

int main()
{
	scanf ("%d", &N);
	for (int i = 1, c; i <= N; i++){
		scanf ("%d", &c);
		I[c].push_back(i);
	}
	for (int i = 1; i < N; i++){
		int x, y; scanf ("%d %d", &x, &y);
		V[x].push_back(y);
		V[y].push_back(x);
	}
	go(1, 0);

	for (int c = 1; c <= N; c++){
		long long ans = 1ll * N * (N + 1) / 2;

		sort(I[c].begin(), I[c].end(), [&](int x, int y){return S[x] > S[y]; });
		vector<pair<int, int> > q;
		for (int x : I[c]){
			int e = 1;
			for (int y : V[x]) if (S[x] < S[y]){
				int z = Z[y] - range(y);
				ans -= 1ll * z * (z + 1) / 2;
				e += z;
			}
			push(S[x], e);
			q.push_back({ S[x], e });
		}

		int z = N - range(1);
		ans -= 1ll * z * (z + 1) / 2;
		for (auto &p : q) push(p.first, -p.second);
		printf ("%lld\n", ans);
	}

	return 0;
}
```