#pragma GCC optimize ("Ofast")
#pragma GCC target ("avx2")
#include<bits/stdc++.h>
#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long ll;
#define rep(i, n) for(int i = 0; i < (n); i++)
#define rep1(i, n) for(int i = 1; i <= (n); i++)
#define co(x) cout << (x) << "\n"
#define cosp(x) cout << (x) << " "
#define ce(x) cerr << (x) << "\n"
#define cesp(x) cerr << (x) << " "
#define pb push_back
#define mp make_pair
#define chmin(x, y) x = min(x, y)
#define chmax(x, y) x = max(x, y)
#define Would
#define you
#define please

int K[200001];
int zentai, N;
ll kotae[200001];
int C[200001];
vector<int> E[200001];

void dfs(int p, int mae) {


	int k = K[C[p]] + zentai;
	int k0 = k;

	for (int to : E[p]) {
		if (mae != to) {
			dfs(to, p);
			int sa = K[C[p]] + zentai - k;
			kotae[C[p]] -= ll(sa) * (sa + 1) / 2;
			k += sa;
		}
	}

	zentai++;
	K[C[p]] = k0 - zentai;
}


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


	cin >> N;
	rep1(i, N) cin >> C[i];
	rep(i, N - 1) {
		int a, b;
		cin >> a >> b;

		E[a].pb(b);
		E[b].pb(a);
	}

	dfs(1, -1);
	ll n = ll(N) * (N + 1) / 2;
	rep1(i, N) {
		int sa = K[i] + zentai;
		co(n + kotae[i] - ll(sa) * (sa + 1) / 2);
	}

	Would you please 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
#pragma GCC optimize ("Ofast")
#pragma GCC target ("avx2")
#include<bits/stdc++.h>
#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long ll;
#define rep(i, n) for(int i = 0; i < (n); i++)
#define rep1(i, n) for(int i = 1; i <= (n); i++)
#define co(x) cout << (x) << "\n"
#define cosp(x) cout << (x) << " "
#define ce(x) cerr << (x) << "\n"
#define cesp(x) cerr << (x) << " "
#define pb push_back
#define mp make_pair
#define chmin(x, y) x = min(x, y)
#define chmax(x, y) x = max(x, y)
#define Would
#define you
#define please

int K[200001];
int zentai, N;
ll kotae[200001];
int C[200001];
vector<int> E[200001];

void dfs(int p, int mae) {


	int k = K[C[p]] + zentai;
	int k0 = k;

	for (int to : E[p]) {
		if (mae != to) {
			dfs(to, p);
			int sa = K[C[p]] + zentai - k;
			kotae[C[p]] -= ll(sa) * (sa + 1) / 2;
			k += sa;
		}
	}

	zentai++;
	K[C[p]] = k0 - zentai;
}


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


	cin >> N;
	rep1(i, N) cin >> C[i];
	rep(i, N - 1) {
		int a, b;
		cin >> a >> b;

		E[a].pb(b);
		E[b].pb(a);
	}

	dfs(1, -1);
	ll n = ll(N) * (N + 1) / 2;
	rep1(i, N) {
		int sa = K[i] + zentai;
		co(n + kotae[i] - ll(sa) * (sa + 1) / 2);
	}

	Would you please return 0;
}
```