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

#define max_N 200010

int color[max_N];
vector<int> edges[max_N];
vector<int> sub_child[max_N];
long sub_size[max_N];
int parent[max_N];
int last_v[max_N];
long no_samecolor_vertex[max_N];
long ans[max_N];

long non_dupcount(int x){
	return (long) x * (x-1) / 2 + x;
}

void dfs(int v, int last = -1){
	parent[v] = last_v[color[v]];
	last_v[color[v]] = v;
	sub_size[v] = 1;
	if(parent[v] != -1) sub_child[parent[v]].push_back(v);
	for(auto x: edges[v]) {
		if(x == last) continue;
		dfs(x, v);
		long t_size = sub_size[x];
		sub_size[v] += t_size;
		for(auto child: sub_child[v])
			t_size -= sub_size[child];
		sub_child[v].clear();
		ans[color[v]] -= non_dupcount(t_size);
	}
	if(parent[v] == -1)
		no_samecolor_vertex[color[v]] -= sub_size[v];
	last_v[color[v]] = parent[v];
}

int main(){
	int n;
	scanf("%d", &n);
	for(int i = 0; i < n; i++) {
		scanf("%d", &color[i]);
		color[i]--;
	}
	for(int i = 0; i < n-1; i++) {
		int x, y;
		scanf("%d%d", &x, &y);
		edges[x-1].push_back(y-1);
		edges[y-1].push_back(x-1);
	}
	for(int i = 0; i < n; i++) {
		no_samecolor_vertex[i] = n;
		last_v[i] = -1;
		ans[i] = non_dupcount(n);
	}
	dfs(0);
	for(int i = 0; i < n; i++) {
		ans[i] -= non_dupcount(no_samecolor_vertex[i]);
		printf("%ld\n", ans[i]);
	}
}

### 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>
using namespace std;

#define max_N 200010

int color[max_N];
vector<int> edges[max_N];
vector<int> sub_child[max_N];
long sub_size[max_N];
int parent[max_N];
int last_v[max_N];
long no_samecolor_vertex[max_N];
long ans[max_N];

long non_dupcount(int x){
	return (long) x * (x-1) / 2 + x;
}

void dfs(int v, int last = -1){
	parent[v] = last_v[color[v]];
	last_v[color[v]] = v;
	sub_size[v] = 1;
	if(parent[v] != -1) sub_child[parent[v]].push_back(v);
	for(auto x: edges[v]) {
		if(x == last) continue;
		dfs(x, v);
		long t_size = sub_size[x];
		sub_size[v] += t_size;
		for(auto child: sub_child[v])
			t_size -= sub_size[child];
		sub_child[v].clear();
		ans[color[v]] -= non_dupcount(t_size);
	}
	if(parent[v] == -1)
		no_samecolor_vertex[color[v]] -= sub_size[v];
	last_v[color[v]] = parent[v];
}

int main(){
	int n;
	scanf("%d", &n);
	for(int i = 0; i < n; i++) {
		scanf("%d", &color[i]);
		color[i]--;
	}
	for(int i = 0; i < n-1; i++) {
		int x, y;
		scanf("%d%d", &x, &y);
		edges[x-1].push_back(y-1);
		edges[y-1].push_back(x-1);
	}
	for(int i = 0; i < n; i++) {
		no_samecolor_vertex[i] = n;
		last_v[i] = -1;
		ans[i] = non_dupcount(n);
	}
	dfs(0);
	for(int i = 0; i < n; i++) {
		ans[i] -= non_dupcount(no_samecolor_vertex[i]);
		printf("%ld\n", ans[i]);
	}
}

```