#include<bits/stdc++.h>
#define ll long long
#define eb emplace_back
using namespace std;
const int N = 2e+5;
vector<ll> v[N], col(N), cnt(N, 0), sz(N, 0), freq(N, 0), ans(N, 0);
void add(ll x, ll y) {
	v[x].eb(y);
	v[y].eb(x);
}
ll nc2(ll n) {
	return( n * (n - 1)) >>1;
}
void dfs(ll nd = 1, ll par = 0) {

	sz[nd] = 1;
	ll count = cnt[col[nd]];
	for (ll ch : v[nd]) {
		if (ch != par) {

			cnt[col[nd]] = 0;

			dfs(ch, nd);

			ans[col[nd]] += nc2(sz[ch] - cnt[col[nd]]);

			sz[nd] += sz[ch];
		}
	}

	cnt[col[nd]] = count + sz[nd];// up subtree +down subtree containing col[nd];


}
int main() {
	ll n;
	cin >> n;
	for (int i = 1; i <= n; i++)cin >> col[i], freq[col[i]]++;
	ll x, y;
	for (int i = 1; i <= n - 1; i++)
		cin >> x >> y,
		add(x, y);
	dfs();
	ll tot = nc2(n);
	for (int i = 1; i <= n; i++)
		ans[i] += nc2(n - cnt[i]);
	for (int i = 1; i <= n; i++)
		cout << tot - ans[i] + freq[i] << "\n";


}



### Prompt

Develop a solution in cpp to the problem described below:

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 eb emplace_back
using namespace std;
const int N = 2e+5;
vector<ll> v[N], col(N), cnt(N, 0), sz(N, 0), freq(N, 0), ans(N, 0);
void add(ll x, ll y) {
	v[x].eb(y);
	v[y].eb(x);
}
ll nc2(ll n) {
	return( n * (n - 1)) >>1;
}
void dfs(ll nd = 1, ll par = 0) {

	sz[nd] = 1;
	ll count = cnt[col[nd]];
	for (ll ch : v[nd]) {
		if (ch != par) {

			cnt[col[nd]] = 0;

			dfs(ch, nd);

			ans[col[nd]] += nc2(sz[ch] - cnt[col[nd]]);

			sz[nd] += sz[ch];
		}
	}

	cnt[col[nd]] = count + sz[nd];// up subtree +down subtree containing col[nd];


}
int main() {
	ll n;
	cin >> n;
	for (int i = 1; i <= n; i++)cin >> col[i], freq[col[i]]++;
	ll x, y;
	for (int i = 1; i <= n - 1; i++)
		cin >> x >> y,
		add(x, y);
	dfs();
	ll tot = nc2(n);
	for (int i = 1; i <= n; i++)
		ans[i] += nc2(n - cnt[i]);
	for (int i = 1; i <= n; i++)
		cout << tot - ans[i] + freq[i] << "\n";


}



```