#include<bits/stdc++.h>
#define ll long long
#define pb push_back
#define all(x) (x).begin(), (x).end()

using namespace std;

const ll maxN = 200005;
ll n;
vector<ll> adj[maxN];
ll c[maxN];
ll sz[maxN];
ll done[maxN];
ll ans[maxN];

void dfs(ll u, ll p) {
	sz[u] = 1;
	ll done_before = done[c[u]];
	for(auto v : adj[u]) {
		if(v != p) {
			done[c[u]] = 0;
			dfs(v, u);
			sz[u] += sz[v];
			ll cnt = sz[v] - done[c[u]];
			ans[c[u]] += cnt*(cnt+1)/2;
			// cout<<u+1<<" -> "<<cnt<<" = "<<sz[v]<<" - "<<done[c[u]]<<endl;
		}
	}
	done[c[u]] = done_before + sz[u];
}

void solve(ll tc) {
	cin>>n;
	for(ll i=0; i<n; ++i) {
		cin>>c[i];
		--c[i];
	}
	for(ll i=0; i<n-1; ++i) {
		ll u, v;
		cin>>u>>v;
		adj[u-1].pb(v-1);
		adj[v-1].pb(u-1);
	}
	dfs(0, -1);
	for(ll i=0; i<n; ++i) {
		ll cnt = sz[0] - done[i];
		ans[i] += cnt*(cnt+1)/2;
		cout<<n*(n+1)/2 - ans[i]<<endl;
	}
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    ll t = 1;
    // cin>>t;
    for(ll tc=1; tc<=t; ++tc) {
        solve(tc);
    }    
    return 0;
}
### Prompt

Your task is to create 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>
#define ll long long
#define pb push_back
#define all(x) (x).begin(), (x).end()

using namespace std;

const ll maxN = 200005;
ll n;
vector<ll> adj[maxN];
ll c[maxN];
ll sz[maxN];
ll done[maxN];
ll ans[maxN];

void dfs(ll u, ll p) {
	sz[u] = 1;
	ll done_before = done[c[u]];
	for(auto v : adj[u]) {
		if(v != p) {
			done[c[u]] = 0;
			dfs(v, u);
			sz[u] += sz[v];
			ll cnt = sz[v] - done[c[u]];
			ans[c[u]] += cnt*(cnt+1)/2;
			// cout<<u+1<<" -> "<<cnt<<" = "<<sz[v]<<" - "<<done[c[u]]<<endl;
		}
	}
	done[c[u]] = done_before + sz[u];
}

void solve(ll tc) {
	cin>>n;
	for(ll i=0; i<n; ++i) {
		cin>>c[i];
		--c[i];
	}
	for(ll i=0; i<n-1; ++i) {
		ll u, v;
		cin>>u>>v;
		adj[u-1].pb(v-1);
		adj[v-1].pb(u-1);
	}
	dfs(0, -1);
	for(ll i=0; i<n; ++i) {
		ll cnt = sz[0] - done[i];
		ans[i] += cnt*(cnt+1)/2;
		cout<<n*(n+1)/2 - ans[i]<<endl;
	}
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    ll t = 1;
    // cin>>t;
    for(ll tc=1; tc<=t; ++tc) {
        solve(tc);
    }    
    return 0;
}
```