#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define mp(a, b) make_pair(a, b)
typedef pair<int, int> pii;
#define ls (rt << 1)
#define rs (rt << 1 | 1) 
typedef pair<int, int> pii; 
const ll mod = 1e9 + 7;
template <class T> T lowbit(T x){return ((x)&(-(x)));}
const int maxn = 2e5 + 10;
int n;
int c[maxn], siz[maxn];
stack<int> stk[maxn]; 
vector<int> E[maxn];
ll val[maxn];
ll calc(ll x) { return x * (x - 1) / 2 + x;}
void dfs(int x, int fa)
{
	siz[x] = 1;
	stk[c[x]].push(x);
	for(auto v : E[x])
		if(v != fa)
		{
			dfs(v, x);
			siz[x] += siz[v];
			int ret = siz[v];
			while(stk[c[x]].top() != x)
			{
				ret -= siz[stk[c[x]].top()];
				stk[c[x]].pop();
			}
			val[c[x]] -= calc(ret);	
		}
}
int main()
{
	scanf("%d", &n);
	for(int i = 1; i <= n; ++i) 
		scanf("%d", &c[i]);
	for(int i = 1; i < n; ++i)
	{
		int u, v;
		scanf("%d%d", &u, &v);
		E[u].push_back(v);
		E[v].push_back(u);
	}
	for(int i = 1; i <= n; ++i) val[i] = calc(n);
	dfs(1, 0);
	for(int i = 1; i <= n; ++i)
	{
		int ret = n;
		while(!stk[i].empty())
		{
			ret -= siz[stk[i].top()];
			stk[i].pop();
		}
		val[i] -= calc(ret);
		cout << val[i] << endl;
	}
	return 0;
}
### Prompt

Please create a solution in cpp 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;
typedef long long ll;
#define mp(a, b) make_pair(a, b)
typedef pair<int, int> pii;
#define ls (rt << 1)
#define rs (rt << 1 | 1) 
typedef pair<int, int> pii; 
const ll mod = 1e9 + 7;
template <class T> T lowbit(T x){return ((x)&(-(x)));}
const int maxn = 2e5 + 10;
int n;
int c[maxn], siz[maxn];
stack<int> stk[maxn]; 
vector<int> E[maxn];
ll val[maxn];
ll calc(ll x) { return x * (x - 1) / 2 + x;}
void dfs(int x, int fa)
{
	siz[x] = 1;
	stk[c[x]].push(x);
	for(auto v : E[x])
		if(v != fa)
		{
			dfs(v, x);
			siz[x] += siz[v];
			int ret = siz[v];
			while(stk[c[x]].top() != x)
			{
				ret -= siz[stk[c[x]].top()];
				stk[c[x]].pop();
			}
			val[c[x]] -= calc(ret);	
		}
}
int main()
{
	scanf("%d", &n);
	for(int i = 1; i <= n; ++i) 
		scanf("%d", &c[i]);
	for(int i = 1; i < n; ++i)
	{
		int u, v;
		scanf("%d%d", &u, &v);
		E[u].push_back(v);
		E[v].push_back(u);
	}
	for(int i = 1; i <= n; ++i) val[i] = calc(n);
	dfs(1, 0);
	for(int i = 1; i <= n; ++i)
	{
		int ret = n;
		while(!stk[i].empty())
		{
			ret -= siz[stk[i].top()];
			stk[i].pop();
		}
		val[i] -= calc(ret);
		cout << val[i] << endl;
	}
	return 0;
}
```