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

const int N = 3e5 + 9;

long long sum[N], ans[N];
vector<int> g[N]; int a[N], sz[N];
inline long long nc2(long long n) {
	return n * (n + 1) / 2;
}
void dfs(int u, int p = 0) {
	sz[u] = 1;
	int c = a[u];
	long long tot = 0;
	for (auto v: g[u]) if (v != p) {
		long long last = sum[c];
		dfs(v, u);
		ans[c] += nc2(sz[v] - (sum[c] - last));
		tot += sum[c] - last;
		sz[u] += sz[v];
	}
	sum[c] -= tot; sum[c] += sz[u];
}
int32_t main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int n; cin >> n;
    for (int i = 1; i <= n; i++) cin >> a[i];
    for (int i = 1; i < n; i++) {
        int u, v; cin >> u >> v;
        g[u].push_back(v);
        g[v].push_back(u);
    }
    dfs(1);
    for (int i = 1; i <= n; i++) ans[i] += nc2(n - sum[i]);
   	for (int i = 1; i <= n; i++) ans[i] = nc2(n) - ans[i];
   	for (int i = 1; i <= n; i++) cout << ans[i] << '\n';
    return 0;
}
### Prompt

Please create a solution in CPP 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;

const int N = 3e5 + 9;

long long sum[N], ans[N];
vector<int> g[N]; int a[N], sz[N];
inline long long nc2(long long n) {
	return n * (n + 1) / 2;
}
void dfs(int u, int p = 0) {
	sz[u] = 1;
	int c = a[u];
	long long tot = 0;
	for (auto v: g[u]) if (v != p) {
		long long last = sum[c];
		dfs(v, u);
		ans[c] += nc2(sz[v] - (sum[c] - last));
		tot += sum[c] - last;
		sz[u] += sz[v];
	}
	sum[c] -= tot; sum[c] += sz[u];
}
int32_t main() {
    ios_base::sync_with_stdio(0);
    cin.tie(0);
    int n; cin >> n;
    for (int i = 1; i <= n; i++) cin >> a[i];
    for (int i = 1; i < n; i++) {
        int u, v; cin >> u >> v;
        g[u].push_back(v);
        g[v].push_back(u);
    }
    dfs(1);
    for (int i = 1; i <= n; i++) ans[i] += nc2(n - sum[i]);
   	for (int i = 1; i <= n; i++) ans[i] = nc2(n) - ans[i];
   	for (int i = 1; i <= n; i++) cout << ans[i] << '\n';
    return 0;
}
```