/** 
  If comrades that you trust gather around you, hope can take physical form and become visible. That’s what I believe.
  */
#include <bits/stdc++.h>
using namespace std;
/**
  (▀̿Ĺ̯▀̿ ̿) (ง ͠° ͟ل͜ ͡°)ง
  */
using ll = long long;
const int N = 2e5 + 7;
vector<int> g[N];
int c[N], cnt[N], sub[N];
ll ans[N];
ll all = 0LL;
int dfs(int v, int p = -1) {
    sub[v] = 1;
    auto bfr = cnt[c[v]];
    for (auto u: g[v]){
        if (u == p) continue;
        ll pre = all - cnt[c[v]];
        sub[v] += dfs(u,v);
        ll nxt = all - cnt[c[v]];
        ans[c[v]] -= (nxt - pre) * (nxt - pre - 1) / 2;
    }
    cnt[c[v]] = bfr + sub[v];
    all++;
    return sub[v];
}

int main(){
	cin.tie(0);
	ios::sync_with_stdio(0);
	int n; cin>>n;
	for (int i = 1;i <= n; i++) { 
        cin >> c[i];
        ans[i] = (ll) n * (n-1) / 2;
    }
	for (int i = 1;i < n; i++){
		int a,b; cin >> a >> b;
		g[a].emplace_back(b);
		g[b].emplace_back(a);
	}
	dfs(1);
	for(int i = 1; i <= n; ++i){
		ll pre = 0, nxt = n - cnt[i];
		ans[i] -= (nxt - pre) * (nxt - pre - 1) / 2;
	}
    for (int i = 1; i <= n; ++i) ans[c[i]]++;
    for (int i = 1; i <= n; ++i) cout << ans[i] << '\n';
	return 0;
}

### Prompt

Please formulate 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
/** 
  If comrades that you trust gather around you, hope can take physical form and become visible. That’s what I believe.
  */
#include <bits/stdc++.h>
using namespace std;
/**
  (▀̿Ĺ̯▀̿ ̿) (ง ͠° ͟ل͜ ͡°)ง
  */
using ll = long long;
const int N = 2e5 + 7;
vector<int> g[N];
int c[N], cnt[N], sub[N];
ll ans[N];
ll all = 0LL;
int dfs(int v, int p = -1) {
    sub[v] = 1;
    auto bfr = cnt[c[v]];
    for (auto u: g[v]){
        if (u == p) continue;
        ll pre = all - cnt[c[v]];
        sub[v] += dfs(u,v);
        ll nxt = all - cnt[c[v]];
        ans[c[v]] -= (nxt - pre) * (nxt - pre - 1) / 2;
    }
    cnt[c[v]] = bfr + sub[v];
    all++;
    return sub[v];
}

int main(){
	cin.tie(0);
	ios::sync_with_stdio(0);
	int n; cin>>n;
	for (int i = 1;i <= n; i++) { 
        cin >> c[i];
        ans[i] = (ll) n * (n-1) / 2;
    }
	for (int i = 1;i < n; i++){
		int a,b; cin >> a >> b;
		g[a].emplace_back(b);
		g[b].emplace_back(a);
	}
	dfs(1);
	for(int i = 1; i <= n; ++i){
		ll pre = 0, nxt = n - cnt[i];
		ans[i] -= (nxt - pre) * (nxt - pre - 1) / 2;
	}
    for (int i = 1; i <= n; ++i) ans[c[i]]++;
    for (int i = 1; i <= n; ++i) cout << ans[i] << '\n';
	return 0;
}

```