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

#define pb push_back
#define fi first
#define SZ(x) (int((x).size()))
#define se second
#define ll long long
#define pq priority_queue
#define MP make_pair
#define pii pair<int,int>
#define mod 998244353
#define inf 0x3f3f3f3f
#define debug(x) cerr<<#x<<"="<<x<<'\n'
#define rep(i, a, b) for (int i=a; i<(b); i++)

const int maxn=2e5+10;
int n;
vector <int> edge[maxn];
vector <int> nxt[maxn];
int c[maxn],lst[maxn],sz[maxn],f[maxn];
ll ans[maxn];

void dfs(int u,int fa) {
	f[u]=fa;
	nxt[lst[c[u]]].pb(u);
	int tmp=lst[c[u]];
	sz[u]=1;
	rep(i,0,SZ(edge[u])) {
		int v=edge[u][i];
		if (v==fa) continue;
		lst[c[u]]=v;
		dfs(v,u);
		sz[u]+=sz[v];
	}
	lst[c[u]]=tmp;
	return;
} 

int main(){
    std::ios::sync_with_stdio(false);
    cin>>n;
    rep(i,1,n+1) cin>>c[i],lst[i]=1;
    rep(i,1,n) {
    	int u,v;
    	cin>>u>>v;
    	edge[u].pb(v);
    	edge[v].pb(u);
	}
	dfs(1,0);
	rep(i,1,n+1) ans[i]=sz[1];
	rep(i,0,SZ(nxt[1])) {
		int cur=nxt[1][i];
		ans[c[cur]]-=sz[cur];
	}
	rep(i,1,n+1) ans[i]=ans[i]+ans[i]*(ans[i]-1)/2;
	rep(i,2,n+1) {
		int tmp=sz[i];
		rep(j,0,SZ(nxt[i])) {
			int v=nxt[i][j];
			tmp-=sz[v];
		}
		ans[c[f[i]]]+=(ll)tmp+(ll)tmp*(ll)(tmp-1)/2ll; 
	}
	rep(i,1,n+1) cout<<(ll)n+(ll)n*(ll)(n-1)/2ll-ans[i]<<endl;
	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
//by szh
#include<bits/stdc++.h>
using namespace std;

#define pb push_back
#define fi first
#define SZ(x) (int((x).size()))
#define se second
#define ll long long
#define pq priority_queue
#define MP make_pair
#define pii pair<int,int>
#define mod 998244353
#define inf 0x3f3f3f3f
#define debug(x) cerr<<#x<<"="<<x<<'\n'
#define rep(i, a, b) for (int i=a; i<(b); i++)

const int maxn=2e5+10;
int n;
vector <int> edge[maxn];
vector <int> nxt[maxn];
int c[maxn],lst[maxn],sz[maxn],f[maxn];
ll ans[maxn];

void dfs(int u,int fa) {
	f[u]=fa;
	nxt[lst[c[u]]].pb(u);
	int tmp=lst[c[u]];
	sz[u]=1;
	rep(i,0,SZ(edge[u])) {
		int v=edge[u][i];
		if (v==fa) continue;
		lst[c[u]]=v;
		dfs(v,u);
		sz[u]+=sz[v];
	}
	lst[c[u]]=tmp;
	return;
} 

int main(){
    std::ios::sync_with_stdio(false);
    cin>>n;
    rep(i,1,n+1) cin>>c[i],lst[i]=1;
    rep(i,1,n) {
    	int u,v;
    	cin>>u>>v;
    	edge[u].pb(v);
    	edge[v].pb(u);
	}
	dfs(1,0);
	rep(i,1,n+1) ans[i]=sz[1];
	rep(i,0,SZ(nxt[1])) {
		int cur=nxt[1][i];
		ans[c[cur]]-=sz[cur];
	}
	rep(i,1,n+1) ans[i]=ans[i]+ans[i]*(ans[i]-1)/2;
	rep(i,2,n+1) {
		int tmp=sz[i];
		rep(j,0,SZ(nxt[i])) {
			int v=nxt[i][j];
			tmp-=sz[v];
		}
		ans[c[f[i]]]+=(ll)tmp+(ll)tmp*(ll)(tmp-1)/2ll; 
	}
	rep(i,1,n+1) cout<<(ll)n+(ll)n*(ll)(n-1)/2ll-ans[i]<<endl;
	return 0;
}

```