#include <cstring>
#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <cmath>
#include <map>
#include<time.h>
using namespace std;
typedef long long ll;
const int maxn=2e5+10;
struct edge {
	int to,Next;
}Edge[maxn<<1];
int Head[maxn],tot;
bool vis[maxn];
ll val[maxn],sum[maxn],sz[maxn],rev[maxn];
void Add(int from, int to) {
	Edge[tot]={to,Head[from]};
	Head[from]=tot++;
	Edge[tot]={from,Head[to]};
	Head[to]=tot++;
}
void dfs(int u, int fa) {
	sz[u]=1;
	ll num=sum[val[u]];
	ll Last=sum[val[u]];
	for (int i=Head[u]; ~i; i=Edge[i].Next) {
		int v=Edge[i].to;
		if (v==fa) continue;
		dfs(v,u);
		sz[u]+=sz[v];
		ll cnt=sz[v]-(sum[val[u]]-Last);
		rev[val[u]]+=(1+cnt)*cnt/2;
		Last=sum[val[u]];
	}
	sum[val[u]]+=sz[u]-(sum[val[u]]-num);
} 
int main(){
	memset(Head,-1,sizeof(Head));
    int n;
    scanf("%d",&n);
    for (int i=1; i<=n; i++) {
    	scanf("%lld",&val[i]);
    	vis[val[i]]=true;
	}
    for (int i=1; i<n; i++) {
    	int u,v;
    	scanf("%d%d",&u,&v);
    	Add(u,v);
	}
    dfs(1,0);
    ll Max=1ll*n*(n+1)/2;
    for (int i=1; i<=n; i++) {
    	if (sum[i]<n&&vis[i]) {
    		ll cnt=n-sum[i];
    		rev[i]+=(cnt+1)*cnt/2;
		}
	}
	for (int i=1; i<=n; i++) printf("%lld\n",vis[i]?Max-rev[i]:0ll);
    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
#include <cstring>
#include <iostream>
#include <stdio.h>
#include <algorithm>
#include <cmath>
#include <map>
#include<time.h>
using namespace std;
typedef long long ll;
const int maxn=2e5+10;
struct edge {
	int to,Next;
}Edge[maxn<<1];
int Head[maxn],tot;
bool vis[maxn];
ll val[maxn],sum[maxn],sz[maxn],rev[maxn];
void Add(int from, int to) {
	Edge[tot]={to,Head[from]};
	Head[from]=tot++;
	Edge[tot]={from,Head[to]};
	Head[to]=tot++;
}
void dfs(int u, int fa) {
	sz[u]=1;
	ll num=sum[val[u]];
	ll Last=sum[val[u]];
	for (int i=Head[u]; ~i; i=Edge[i].Next) {
		int v=Edge[i].to;
		if (v==fa) continue;
		dfs(v,u);
		sz[u]+=sz[v];
		ll cnt=sz[v]-(sum[val[u]]-Last);
		rev[val[u]]+=(1+cnt)*cnt/2;
		Last=sum[val[u]];
	}
	sum[val[u]]+=sz[u]-(sum[val[u]]-num);
} 
int main(){
	memset(Head,-1,sizeof(Head));
    int n;
    scanf("%d",&n);
    for (int i=1; i<=n; i++) {
    	scanf("%lld",&val[i]);
    	vis[val[i]]=true;
	}
    for (int i=1; i<n; i++) {
    	int u,v;
    	scanf("%d%d",&u,&v);
    	Add(u,v);
	}
    dfs(1,0);
    ll Max=1ll*n*(n+1)/2;
    for (int i=1; i<=n; i++) {
    	if (sum[i]<n&&vis[i]) {
    		ll cnt=n-sum[i];
    		rev[i]+=(cnt+1)*cnt/2;
		}
	}
	for (int i=1; i<=n; i++) printf("%lld\n",vis[i]?Max-rev[i]:0ll);
    return 0;
}

```