#include <bits/stdc++.h>
#define int long long
#define N (int) 3e5
using namespace std;
int n;
int col[N];
int num[N];	
struct Edge {
	int v, nxt;
} edge[N<<1];
int fst[N], ecnt;
void add_edge (int u, int v) {
	edge[ecnt].v = v, edge[ecnt].nxt = fst[u], fst[u] = ecnt ++;
}
void init_e () {
	memset (fst, -1, sizeof (fst));
}

int cnt[N], ans[N], sz[N];
void dfs (int u, int f) {
	sz[u] = 1;
	int org = cnt[col[u]];
	for (int e = fst[u]; e != -1; e = edge[e].nxt) {
		int v = edge[e].v;
		if (v == f) continue;
		int tmp = cnt[col[u]];
		dfs (v, u);
		int tmpp = sz[v] - (cnt[col[u]] - tmp);
		ans[col[u]] += (tmpp - 1) * tmpp >> 1;
		sz[u] += sz[v];
	}
	
	cnt[col[u]] = org + sz[u];
}
signed main () {
	init_e ();
	
	cin >> n;
	for (int i = 1; i <= n; i ++) cin >> col[i], num[col[i]] ++;
	for (int i = 1; i < n; i ++) {
		int u, v; cin >> u >> v;
		add_edge (u, v), add_edge (v, u);
	}
	
	dfs (1, 0);
	
	int tot = n * (n-1) >> 1;
	for (int i = 1; i <= n; i ++) {
		int tmp = n - cnt[i];
		if (! tmp) continue;
		ans[i] += tmp * (tmp - 1) >> 1;
	}
	for (int i = 1; i <= n; i ++) cout << (tot - ans[i] + num[i]) << endl;
}
### Prompt

Generate 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 int long long
#define N (int) 3e5
using namespace std;
int n;
int col[N];
int num[N];	
struct Edge {
	int v, nxt;
} edge[N<<1];
int fst[N], ecnt;
void add_edge (int u, int v) {
	edge[ecnt].v = v, edge[ecnt].nxt = fst[u], fst[u] = ecnt ++;
}
void init_e () {
	memset (fst, -1, sizeof (fst));
}

int cnt[N], ans[N], sz[N];
void dfs (int u, int f) {
	sz[u] = 1;
	int org = cnt[col[u]];
	for (int e = fst[u]; e != -1; e = edge[e].nxt) {
		int v = edge[e].v;
		if (v == f) continue;
		int tmp = cnt[col[u]];
		dfs (v, u);
		int tmpp = sz[v] - (cnt[col[u]] - tmp);
		ans[col[u]] += (tmpp - 1) * tmpp >> 1;
		sz[u] += sz[v];
	}
	
	cnt[col[u]] = org + sz[u];
}
signed main () {
	init_e ();
	
	cin >> n;
	for (int i = 1; i <= n; i ++) cin >> col[i], num[col[i]] ++;
	for (int i = 1; i < n; i ++) {
		int u, v; cin >> u >> v;
		add_edge (u, v), add_edge (v, u);
	}
	
	dfs (1, 0);
	
	int tot = n * (n-1) >> 1;
	for (int i = 1; i <= n; i ++) {
		int tmp = n - cnt[i];
		if (! tmp) continue;
		ans[i] += tmp * (tmp - 1) >> 1;
	}
	for (int i = 1; i <= n; i ++) cout << (tot - ans[i] + num[i]) << endl;
}
```