#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
void read(int &x){
	scanf("%d",&x);
}
void write(ll x){
	if (x>9) write(x/10);
	putchar(x%10+'0'); 
}
int n,a[200004],sub[200004];
vector<int>g[200004],v[200004];
ll ans[200004];
int ans1[200004],res[200004];
void dfs(int x,int par){
	sub[x]=1;
	for (int i=0;i<g[x].size();i++)if(g[x][i]!=par)
		dfs(g[x][i],x),sub[x]+=sub[g[x][i]];
}
void dfs1(int x,int par){
	if (v[a[x]].empty()) ans1[a[x]]-=sub[x];
	else res[v[a[x]][v[a[x]].size()-1]]-=sub[x];
	for (int i=0;i<g[x].size();i++) if (g[x][i]!=par){
		v[a[x]].push_back(g[x][i]);
		dfs1(g[x][i],x);
		ans[a[x]]+=1ll*res[g[x][i]]*(res[g[x][i]]+1)/2;
		v[a[x]].pop_back(); 
	}
}
int x,y;
ll A;
int main(){
	read(n),A=1ll*n*(n+1)/2;
	for (int i=1;i<=n;i++) read(a[i]),ans1[i]=n;
	for (int i=1;i<n;i++)read(x),read(y),g[x].push_back(y),g[y].push_back(x); 
	dfs(1,-1);
	for (int i=1;i<=n;i++) res[i]=sub[i];
	dfs1(1,-1);
	for (int i=1;i<=n;i++){
		ll res=A-ans[i]-1ll*ans1[i]*(ans1[i]+1)/2;
		write(res),puts("");
	}
}
### 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>
using namespace std;
typedef long long ll;
void read(int &x){
	scanf("%d",&x);
}
void write(ll x){
	if (x>9) write(x/10);
	putchar(x%10+'0'); 
}
int n,a[200004],sub[200004];
vector<int>g[200004],v[200004];
ll ans[200004];
int ans1[200004],res[200004];
void dfs(int x,int par){
	sub[x]=1;
	for (int i=0;i<g[x].size();i++)if(g[x][i]!=par)
		dfs(g[x][i],x),sub[x]+=sub[g[x][i]];
}
void dfs1(int x,int par){
	if (v[a[x]].empty()) ans1[a[x]]-=sub[x];
	else res[v[a[x]][v[a[x]].size()-1]]-=sub[x];
	for (int i=0;i<g[x].size();i++) if (g[x][i]!=par){
		v[a[x]].push_back(g[x][i]);
		dfs1(g[x][i],x);
		ans[a[x]]+=1ll*res[g[x][i]]*(res[g[x][i]]+1)/2;
		v[a[x]].pop_back(); 
	}
}
int x,y;
ll A;
int main(){
	read(n),A=1ll*n*(n+1)/2;
	for (int i=1;i<=n;i++) read(a[i]),ans1[i]=n;
	for (int i=1;i<n;i++)read(x),read(y),g[x].push_back(y),g[y].push_back(x); 
	dfs(1,-1);
	for (int i=1;i<=n;i++) res[i]=sub[i];
	dfs1(1,-1);
	for (int i=1;i<=n;i++){
		ll res=A-ans[i]-1ll*ans1[i]*(ans1[i]+1)/2;
		write(res),puts("");
	}
}
```