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

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

void go(int x, int l)
{
	S[x] = ++K;

	auto &i = I[C[x]];
	for (auto &y : V[x]) if (y != l){
		go(y, x);
		int n = Z[y];
		while (!i.empty()){
			int z = i.back();
			if (S[y] <= S[z]){
				n -= Z[z];
				i.pop_back();
			}
			else break;
		}
		A[C[x]] -= 1ll * n * (n + 1) / 2;
	}
	i.push_back(x);

	E[x] = K;
	Z[x] = E[x] - S[x] + 1;
}

int main()
{
	scanf ("%d", &N);
	for (int i = 1; i <= N; i++) scanf ("%d", &C[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++){
		A[c] += 1ll * N * (N + 1) / 2;
		int n = N;
		for (auto &x : I[c]) n -= Z[x];
		A[c] -= 1ll * n * (n + 1) / 2;
		printf ("%lld\n", A[c]);
	}

	return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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 <vector>
using namespace std;

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

void go(int x, int l)
{
	S[x] = ++K;

	auto &i = I[C[x]];
	for (auto &y : V[x]) if (y != l){
		go(y, x);
		int n = Z[y];
		while (!i.empty()){
			int z = i.back();
			if (S[y] <= S[z]){
				n -= Z[z];
				i.pop_back();
			}
			else break;
		}
		A[C[x]] -= 1ll * n * (n + 1) / 2;
	}
	i.push_back(x);

	E[x] = K;
	Z[x] = E[x] - S[x] + 1;
}

int main()
{
	scanf ("%d", &N);
	for (int i = 1; i <= N; i++) scanf ("%d", &C[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++){
		A[c] += 1ll * N * (N + 1) / 2;
		int n = N;
		for (auto &x : I[c]) n -= Z[x];
		A[c] -= 1ll * n * (n + 1) / 2;
		printf ("%lld\n", A[c]);
	}

	return 0;
}

```