#include <bits/stdc++.h>
using namespace std;

template<typename T>
void out(T x) { cout << x << endl; exit(0); }
#define watch(x) cout << (#x) << " is " << (x) << endl





using ll = long long;


const int maxn = 1e6 + 5;


int n;
int c[maxn];
vector<int> g[maxn];
int siz[maxn];


ll ans[maxn];

void dfs(int at, int p, map<ll,ll>& mp) {
    siz[at]=1;
    int C = c[at];
    for (int to: g[at]) {
	if (to == p) continue;
	map<ll,ll> mp_to;
	dfs(to, at, mp_to);
	siz[at]+=siz[to];
	ll middle = siz[to]-mp_to[C];
	ans[C] -= (middle*(middle+1)/2);
	if (mp.size() < mp_to.size()) {
	    swap(mp, mp_to);
	}
	for (auto pr: mp_to) {
	    mp[pr.first] += pr.second;
	}
    }
    
    mp[C] = siz[at];
}

int main() {
    ios_base::sync_with_stdio(false); cin.tie(0);  cout.tie(0);

    
    cin>>n;
    for (int i=1; i<=n; i++) {
	cin>>c[i];
    }

    for (int i=0; i<n-1; i++) {
	int u,v; cin>>u>>v;
	g[u].push_back(v);
	g[v].push_back(u);
    }

    for (int i=1; i<=n; i++) {
	ans[i]=1ll*n*(n+1)/2;
    }

    map<ll,ll> mp;
    dfs(1,0,mp);

    for (int i=1; i<=n; i++) {
	ll top = n-mp[i];
	ans[i] -= (top*(top+1)/2);
    }

    for (int i=1; i<=n; i++) {
	cout<<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
#include <bits/stdc++.h>
using namespace std;

template<typename T>
void out(T x) { cout << x << endl; exit(0); }
#define watch(x) cout << (#x) << " is " << (x) << endl





using ll = long long;


const int maxn = 1e6 + 5;


int n;
int c[maxn];
vector<int> g[maxn];
int siz[maxn];


ll ans[maxn];

void dfs(int at, int p, map<ll,ll>& mp) {
    siz[at]=1;
    int C = c[at];
    for (int to: g[at]) {
	if (to == p) continue;
	map<ll,ll> mp_to;
	dfs(to, at, mp_to);
	siz[at]+=siz[to];
	ll middle = siz[to]-mp_to[C];
	ans[C] -= (middle*(middle+1)/2);
	if (mp.size() < mp_to.size()) {
	    swap(mp, mp_to);
	}
	for (auto pr: mp_to) {
	    mp[pr.first] += pr.second;
	}
    }
    
    mp[C] = siz[at];
}

int main() {
    ios_base::sync_with_stdio(false); cin.tie(0);  cout.tie(0);

    
    cin>>n;
    for (int i=1; i<=n; i++) {
	cin>>c[i];
    }

    for (int i=0; i<n-1; i++) {
	int u,v; cin>>u>>v;
	g[u].push_back(v);
	g[v].push_back(u);
    }

    for (int i=1; i<=n; i++) {
	ans[i]=1ll*n*(n+1)/2;
    }

    map<ll,ll> mp;
    dfs(1,0,mp);

    for (int i=1; i<=n; i++) {
	ll top = n-mp[i];
	ans[i] -= (top*(top+1)/2);
    }

    for (int i=1; i<=n; i++) {
	cout<<ans[i]<<"\n";
    }
    
    
    return 0;
}

```