#include<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#define size sz
using namespace std;
typedef long long ll;

const int INF = 0x3f3f3f3f;
const int MAXN=200005;

int n,len;
int color[MAXN],head[MAXN],size[MAXN],num[MAXN];
ll ans[MAXN];

struct node{
	int to,next;
} e[MAXN * 2];

void add(int u,int v){
	e[++len].to = v;
	e[len].next = head[u];
	head[u] = len;
}

ll f(ll n){
	return 1ll * n * (n + 1) / 2;
}

void dfs(int u,int fa){	
	int c = color[u];
	int lst,d;
	size[u] = 1;
	ll tmp = num[c];
	for(int i = head[u];i != -1;i = e[i].next){
		int v = e[i].to;
		if(v == fa)
			continue;
		lst = num[c];
		dfs(v,u);
		d = num[c] - lst;
		ans[c] -= f(size[v] - d);
		size[u] += size[v];
	}
	num[c] = tmp + size[u];
}

int main(){
	memset(head,-1,sizeof(head));
	scanf("%d",&n);
	for(int i = 1;i <= n;i++){
		scanf("%d",&color[i]);
		ans[i] = f(n);
	}
	for(int i = 1;i < n;i++){
		int u,v;
		scanf("%d%d",&u,&v);
		add(u,v);
		add(v,u);
	}
	dfs(1,0);
	for(int i = 1;i <= n;i++){
		ans[i] -= f(n - num[i]);
		printf("%lld\n",ans[i]);
	}
	return 0;
}
### Prompt

Your task is to create 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<iostream>
#include<cstring>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#define size sz
using namespace std;
typedef long long ll;

const int INF = 0x3f3f3f3f;
const int MAXN=200005;

int n,len;
int color[MAXN],head[MAXN],size[MAXN],num[MAXN];
ll ans[MAXN];

struct node{
	int to,next;
} e[MAXN * 2];

void add(int u,int v){
	e[++len].to = v;
	e[len].next = head[u];
	head[u] = len;
}

ll f(ll n){
	return 1ll * n * (n + 1) / 2;
}

void dfs(int u,int fa){	
	int c = color[u];
	int lst,d;
	size[u] = 1;
	ll tmp = num[c];
	for(int i = head[u];i != -1;i = e[i].next){
		int v = e[i].to;
		if(v == fa)
			continue;
		lst = num[c];
		dfs(v,u);
		d = num[c] - lst;
		ans[c] -= f(size[v] - d);
		size[u] += size[v];
	}
	num[c] = tmp + size[u];
}

int main(){
	memset(head,-1,sizeof(head));
	scanf("%d",&n);
	for(int i = 1;i <= n;i++){
		scanf("%d",&color[i]);
		ans[i] = f(n);
	}
	for(int i = 1;i < n;i++){
		int u,v;
		scanf("%d%d",&u,&v);
		add(u,v);
		add(v,u);
	}
	dfs(1,0);
	for(int i = 1;i <= n;i++){
		ans[i] -= f(n - num[i]);
		printf("%lld\n",ans[i]);
	}
	return 0;
}
```