#include <bits/stdc++.h>
#define int long long

using namespace std;
const int N = 3e5;

int n;
int size[N], color[N], col[N], res[N];
vector < int > a[N];

int sum(int x) {
	return x * (x+1) / 2;
}

int read(void)
{
	int s = 0, w = 0; char c = getchar();
	while (c < '0' || c > '9') w |= c == '-', c = getchar();
	while (c >= '0' && c <= '9') s = s*10+c-48, c = getchar();
	return w ? -s : s;
}

void dfs(int x,int fa)
{
	size[x] = 1;
	int temp = col[color[x]], c = color[x];
	for (int i=0;i<a[x].size();++i)
	{
		int y = a[x][i];
		if (y == fa) continue;
		int t = col[c];
		dfs(y,x);
		int Ad = col[c] - t;
		size[x] += size[y];
		res[c] -= sum(size[y] - Ad);
	}
	col[c] = temp + size[x];
	return;
}

signed main(void)
{
	n = read();
	for (int i=1;i<=n;++i) color[i] = read();
	for (int i=1,x,y;i<n;++i)
	{
		x = read(), y = read();
		a[x].push_back(y);
		a[y].push_back(x);
	}
	for (int i=1;i<=n;++i) res[i] = sum(n);
	dfs(1,0);
	for (int i=1;i<=n;++i) 
		printf("%lld\n", res[i] - sum(n-col[i]));
	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 <bits/stdc++.h>
#define int long long

using namespace std;
const int N = 3e5;

int n;
int size[N], color[N], col[N], res[N];
vector < int > a[N];

int sum(int x) {
	return x * (x+1) / 2;
}

int read(void)
{
	int s = 0, w = 0; char c = getchar();
	while (c < '0' || c > '9') w |= c == '-', c = getchar();
	while (c >= '0' && c <= '9') s = s*10+c-48, c = getchar();
	return w ? -s : s;
}

void dfs(int x,int fa)
{
	size[x] = 1;
	int temp = col[color[x]], c = color[x];
	for (int i=0;i<a[x].size();++i)
	{
		int y = a[x][i];
		if (y == fa) continue;
		int t = col[c];
		dfs(y,x);
		int Ad = col[c] - t;
		size[x] += size[y];
		res[c] -= sum(size[y] - Ad);
	}
	col[c] = temp + size[x];
	return;
}

signed main(void)
{
	n = read();
	for (int i=1;i<=n;++i) color[i] = read();
	for (int i=1,x,y;i<n;++i)
	{
		x = read(), y = read();
		a[x].push_back(y);
		a[y].push_back(x);
	}
	for (int i=1;i<=n;++i) res[i] = sum(n);
	dfs(1,0);
	for (int i=1;i<=n;++i) 
		printf("%lld\n", res[i] - sum(n-col[i]));
	return 0;
}
```