//In the name of GOD
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;

const ll maxn = 2e5 + 100;
const ll mod = 1e9 + 7;
const ll inf = 1e18;

#define fast_io ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define file_io freopen("input.txt" , "r+" , stdin) ; freopen("output.txt" , "w+" , stdout);
#define pb push_back
#define Mp make_pair
#define pll pair<ll, ll>
#define F first
#define S second

ll n, ans[maxn], c[maxn], sz[maxn];
unordered_map<ll, ll> cnt[maxn];
vector<ll> g[maxn];
bool vis[maxn];

void dfs(ll v){
	ll u;
	vis[v] = 1;
	sz[v] = 1;
	for(ll i = 0; i < g[v].size(); i++){
		u = g[v][i];
		if(vis[u]) continue;
		dfs(u);
		sz[v] += sz[u];
		ll x = sz[u] - cnt[u][c[v]];
		ans[c[v]] += x * (x + 1) / 2;
		if(cnt[v].size() < cnt[u].size()) cnt[v].swap(cnt[u]);
		for(unordered_map<ll, ll>::iterator it = cnt[u].begin(); it != cnt[u].end(); it++){
			cnt[v][it -> F] += it -> S;
		}
	}
	cnt[v][c[v]] = sz[v];
	return;
}

int main(){
	ll v, u;
	fast_io;
	
	cin >> n;
	for(ll i = 1; i <= n; i++){
		cin >> c[i];
	}
	for(ll i = 1; i < n; i++){
		cin >> v >> u;
		g[v].pb(u);
		g[u].pb(v);
	}
	dfs(1);
	for(ll i = 1; i <= n; i++){
		ll x = n - cnt[1][i];
		ans[i] += x * (x + 1) / 2;
		cout << n * (n + 1) / 2 - ans[i] << '\n';
	}
	
	return 0;
}

### Prompt

Generate 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
//In the name of GOD
#include <bits/stdc++.h>
using namespace std;

typedef long long ll;

const ll maxn = 2e5 + 100;
const ll mod = 1e9 + 7;
const ll inf = 1e18;

#define fast_io ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define file_io freopen("input.txt" , "r+" , stdin) ; freopen("output.txt" , "w+" , stdout);
#define pb push_back
#define Mp make_pair
#define pll pair<ll, ll>
#define F first
#define S second

ll n, ans[maxn], c[maxn], sz[maxn];
unordered_map<ll, ll> cnt[maxn];
vector<ll> g[maxn];
bool vis[maxn];

void dfs(ll v){
	ll u;
	vis[v] = 1;
	sz[v] = 1;
	for(ll i = 0; i < g[v].size(); i++){
		u = g[v][i];
		if(vis[u]) continue;
		dfs(u);
		sz[v] += sz[u];
		ll x = sz[u] - cnt[u][c[v]];
		ans[c[v]] += x * (x + 1) / 2;
		if(cnt[v].size() < cnt[u].size()) cnt[v].swap(cnt[u]);
		for(unordered_map<ll, ll>::iterator it = cnt[u].begin(); it != cnt[u].end(); it++){
			cnt[v][it -> F] += it -> S;
		}
	}
	cnt[v][c[v]] = sz[v];
	return;
}

int main(){
	ll v, u;
	fast_io;
	
	cin >> n;
	for(ll i = 1; i <= n; i++){
		cin >> c[i];
	}
	for(ll i = 1; i < n; i++){
		cin >> v >> u;
		g[v].pb(u);
		g[u].pb(v);
	}
	dfs(1);
	for(ll i = 1; i <= n; i++){
		ll x = n - cnt[1][i];
		ans[i] += x * (x + 1) / 2;
		cout << n * (n + 1) / 2 - ans[i] << '\n';
	}
	
	return 0;
}

```