#include<bits/stdc++.h>
#define sz(x) ((int)x.size())
#define pb push_back
#define ii pair<int,int>
#define ppb pop_back
#define orta ((bas+son)>>1)
#define st first
#define nd second
#define ll long long
#define N 200005
#define inf 100000000000000ll
#define MOD 998244353
#define LOG 31
#define EPS 0.000000001
#define M 305
#define PI 3.14159265359
using namespace std;

int n, t;
int be[N], sub[N], c[N];
ll ans[N];
vector<int> has[N], v[N];

void evaluate(int node, int color) {

	int cnt = sub[node];
	while(sz(has[color]) and be[has[color].back()] >= be[node]) {
		cnt -= sub[has[color].back()];
		has[color].pop_back();
	}
	ans[color] -= (ll) cnt * (cnt + 1) / 2;

}

void dfs(int node, int dad) {

	be[node] = ++t;
	sub[node] = 1;
	for(int go: v[node]) {
		if(go == dad) continue ;
		dfs(go, node);
		sub[node] += sub[go];
	}
	has[c[node]].pb(node);
	if(!dad) 
		for(int i = 1; i <= n; i++)
			evaluate(node, i);
	else
		evaluate(node, c[dad]);

}

int main() {

	ios_base::sync_with_stdio(0);
	cin.tie(0);cout.tie(0);

	cin >> n;
	for(int i = 1; i <= n ; i++) 
		cin >> c[i];
	for(int i = 1; i < n; i++) {
		int x, y;
		cin >> x >> y;
		v[x].pb(y);
		v[y].pb(x);
	}
	dfs(1, 0);
	for(int i = 1; i <= n; i++) 
		cout << (ll) n * (n + 1) / 2 + ans[i] << "\n";

}
### Prompt

Develop a solution in Cpp 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>
#define sz(x) ((int)x.size())
#define pb push_back
#define ii pair<int,int>
#define ppb pop_back
#define orta ((bas+son)>>1)
#define st first
#define nd second
#define ll long long
#define N 200005
#define inf 100000000000000ll
#define MOD 998244353
#define LOG 31
#define EPS 0.000000001
#define M 305
#define PI 3.14159265359
using namespace std;

int n, t;
int be[N], sub[N], c[N];
ll ans[N];
vector<int> has[N], v[N];

void evaluate(int node, int color) {

	int cnt = sub[node];
	while(sz(has[color]) and be[has[color].back()] >= be[node]) {
		cnt -= sub[has[color].back()];
		has[color].pop_back();
	}
	ans[color] -= (ll) cnt * (cnt + 1) / 2;

}

void dfs(int node, int dad) {

	be[node] = ++t;
	sub[node] = 1;
	for(int go: v[node]) {
		if(go == dad) continue ;
		dfs(go, node);
		sub[node] += sub[go];
	}
	has[c[node]].pb(node);
	if(!dad) 
		for(int i = 1; i <= n; i++)
			evaluate(node, i);
	else
		evaluate(node, c[dad]);

}

int main() {

	ios_base::sync_with_stdio(0);
	cin.tie(0);cout.tie(0);

	cin >> n;
	for(int i = 1; i <= n ; i++) 
		cin >> c[i];
	for(int i = 1; i < n; i++) {
		int x, y;
		cin >> x >> y;
		v[x].pb(y);
		v[y].pb(x);
	}
	dfs(1, 0);
	for(int i = 1; i <= n; i++) 
		cout << (ll) n * (n + 1) / 2 + ans[i] << "\n";

}
```