#include <bits/stdc++.h>
using namespace std;
#define rep(i,n) for(int i=0; i<(n); i++)
using ll = long long;
constexpr int MX = 200005;
int n, c[MX], a, b, mu[MX*2];
vector<int> G[MX];
stack<int> par[MX];
ll ans[MX];
int count(int u, int p){
	int ret = 1, t, premu = 0;
	par[c[u]].push(u);
	for(int nx: G[u]) if(nx != p){
		t = count(nx, u);
		ret += t;
		t += mu[u] - premu;
		ans[c[u]] -= (ll)t * (t+1) / 2;
		premu = mu[u];
	}
	par[c[u]].pop();
	mu[par[c[u]].top()] -= ret;
	return ret;
}
int main(){
	scanf("%d", &n);
	rep(i,n){
		scanf("%d", c+i);
		c[i]--;
	}
	rep(i,n-1){
		scanf("%d%d", &a, &b);
		a--; b--;
		G[a].push_back(b);
		G[b].push_back(a);
	}
	rep(i,n) par[i].push(n+i);
	fill(ans, ans+n, (ll)n * (n+1) / 2);
	fill(mu+n, mu+n*2, n);
	count(0,n);
	rep(i,n) printf("%lld\n", ans[i] - (ll)mu[i+n] * (mu[i+n]+1) / 2);
}
### Prompt

Please provide a cpp coded solution to the problem described below:

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 rep(i,n) for(int i=0; i<(n); i++)
using ll = long long;
constexpr int MX = 200005;
int n, c[MX], a, b, mu[MX*2];
vector<int> G[MX];
stack<int> par[MX];
ll ans[MX];
int count(int u, int p){
	int ret = 1, t, premu = 0;
	par[c[u]].push(u);
	for(int nx: G[u]) if(nx != p){
		t = count(nx, u);
		ret += t;
		t += mu[u] - premu;
		ans[c[u]] -= (ll)t * (t+1) / 2;
		premu = mu[u];
	}
	par[c[u]].pop();
	mu[par[c[u]].top()] -= ret;
	return ret;
}
int main(){
	scanf("%d", &n);
	rep(i,n){
		scanf("%d", c+i);
		c[i]--;
	}
	rep(i,n-1){
		scanf("%d%d", &a, &b);
		a--; b--;
		G[a].push_back(b);
		G[b].push_back(a);
	}
	rep(i,n) par[i].push(n+i);
	fill(ans, ans+n, (ll)n * (n+1) / 2);
	fill(mu+n, mu+n*2, n);
	count(0,n);
	rep(i,n) printf("%lld\n", ans[i] - (ll)mu[i+n] * (mu[i+n]+1) / 2);
}
```