#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
#define endl '\n'
#define all(x) (x).begin(),(x).end()
const int INF=1e9+5;
const int N=2e5+5;
const ll oo=1e18+5;
const ll mod=1e9+7;

vector<int> v[N];
vector<int> sizee;
vector<int> cnt;
vector<int> c;
vector<map<int,int>> dp;
vector<vector<ll>> res;

void dfs(int u,int par){
	for(int x:v[u]){
		if(x==par) continue;
		dfs(x,u);
		cnt[u]+=cnt[x];
		int tmpcnt=cnt[x];
		if(dp[x].count(c[x])) tmpcnt-=dp[x][c[u]];
		res[c[u]].push_back(tmpcnt);
		if((int)dp[u].size()<(int)dp[x].size()) swap(dp[u],dp[x]);
		for(auto& xx:dp[x]) dp[u][xx.first]+=xx.second;
	}
	++cnt[u];
	dp[u][c[u]]=cnt[u];
}

int main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	int n;
	cin>>n;
	sizee.resize(n);
	cnt.resize(n);
	c.resize(n);
	res.resize(n);
	dp.resize(n);
	for(int& x:c) cin>>x,--x;
	for(int i=0;i<n-1;++i){
		int x,y;
		cin>>x>>y;
		--x;
		--y;
		v[x].push_back(y);
		v[y].push_back(x);
	}
	dfs(0,-1);
	for(int i=0;i<n;++i) res[i].push_back(n-dp[0][i]);
	vector<ll> ans(n,1ll*n*(n+1)/2);
	for(int i=0;i<n;++i){
		for(ll x:res[i]) ans[i]-=x*(x+1)/2;
		cout<<ans[i]<<" ";
	}
	return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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 endl '\n'
#define all(x) (x).begin(),(x).end()
const int INF=1e9+5;
const int N=2e5+5;
const ll oo=1e18+5;
const ll mod=1e9+7;

vector<int> v[N];
vector<int> sizee;
vector<int> cnt;
vector<int> c;
vector<map<int,int>> dp;
vector<vector<ll>> res;

void dfs(int u,int par){
	for(int x:v[u]){
		if(x==par) continue;
		dfs(x,u);
		cnt[u]+=cnt[x];
		int tmpcnt=cnt[x];
		if(dp[x].count(c[x])) tmpcnt-=dp[x][c[u]];
		res[c[u]].push_back(tmpcnt);
		if((int)dp[u].size()<(int)dp[x].size()) swap(dp[u],dp[x]);
		for(auto& xx:dp[x]) dp[u][xx.first]+=xx.second;
	}
	++cnt[u];
	dp[u][c[u]]=cnt[u];
}

int main(){
	ios::sync_with_stdio(0);
	cin.tie(0);
	int n;
	cin>>n;
	sizee.resize(n);
	cnt.resize(n);
	c.resize(n);
	res.resize(n);
	dp.resize(n);
	for(int& x:c) cin>>x,--x;
	for(int i=0;i<n-1;++i){
		int x,y;
		cin>>x>>y;
		--x;
		--y;
		v[x].push_back(y);
		v[y].push_back(x);
	}
	dfs(0,-1);
	for(int i=0;i<n;++i) res[i].push_back(n-dp[0][i]);
	vector<ll> ans(n,1ll*n*(n+1)/2);
	for(int i=0;i<n;++i){
		for(ll x:res[i]) ans[i]-=x*(x+1)/2;
		cout<<ans[i]<<" ";
	}
	return 0;
}

```