#include <bits/stdc++.h>
#define MOD 1000000007LL
using namespace std;
typedef long long ll;
typedef pair<int,int> P;

int n;
int c[200005];
vector<int> G[200005];
ll ans[200005];
ll cnt[200005];
ll val[200005];
ll all=0;

int dfs(int v,int p){
	int su=1;
	ll tmp=val[c[v]];
	for(int i=0;i<G[v].size();i++){
		int nv=G[v][i];
		if(nv==p)continue;
		ll prv=all-val[c[v]];
		su+=dfs(nv,v);
		ll nxt=all-val[c[v]];
		ans[c[v]]-=(nxt-prv)*(nxt-prv-1)/2LL;
	}
	all++;
	val[c[v]]=tmp+su;
	return (cnt[v]=su);
}

int main(void){
	scanf("%d",&n);
	for(int i=0;i<n;i++){
		scanf("%d",&c[i]);
		c[i]--;
	}
	for(int i=0;i<n-1;i++){
		int a,b;
		scanf("%d%d",&a,&b);
		a--;
		b--;
		G[a].push_back(b);
		G[b].push_back(a);
	}
	for(int i=0;i<n;i++){
		ans[i]=(ll)n*(n-1)/2LL;
	}
	dfs(0,-1);
	for(int i=0;i<n;i++){
		ll prv=0,nxt=n-val[i];
		ans[i]-=(nxt-prv)*(nxt-prv-1)/2LL;
	}
	for(int i=0;i<n;i++){
		ans[c[i]]++;
	}
	for(int i=0;i<n;i++){
		printf("%lld\n",ans[i]);
	}
	return 0;
}

### Prompt

Create a solution in Cpp for 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 MOD 1000000007LL
using namespace std;
typedef long long ll;
typedef pair<int,int> P;

int n;
int c[200005];
vector<int> G[200005];
ll ans[200005];
ll cnt[200005];
ll val[200005];
ll all=0;

int dfs(int v,int p){
	int su=1;
	ll tmp=val[c[v]];
	for(int i=0;i<G[v].size();i++){
		int nv=G[v][i];
		if(nv==p)continue;
		ll prv=all-val[c[v]];
		su+=dfs(nv,v);
		ll nxt=all-val[c[v]];
		ans[c[v]]-=(nxt-prv)*(nxt-prv-1)/2LL;
	}
	all++;
	val[c[v]]=tmp+su;
	return (cnt[v]=su);
}

int main(void){
	scanf("%d",&n);
	for(int i=0;i<n;i++){
		scanf("%d",&c[i]);
		c[i]--;
	}
	for(int i=0;i<n-1;i++){
		int a,b;
		scanf("%d%d",&a,&b);
		a--;
		b--;
		G[a].push_back(b);
		G[b].push_back(a);
	}
	for(int i=0;i<n;i++){
		ans[i]=(ll)n*(n-1)/2LL;
	}
	dfs(0,-1);
	for(int i=0;i<n;i++){
		ll prv=0,nxt=n-val[i];
		ans[i]-=(nxt-prv)*(nxt-prv-1)/2LL;
	}
	for(int i=0;i<n;i++){
		ans[c[i]]++;
	}
	for(int i=0;i<n;i++){
		printf("%lld\n",ans[i]);
	}
	return 0;
}

```