#include<cstdio>
#include<cstring>
#include<cmath>
#include<cassert>
#include<iostream>
#include<iomanip>
#include<map>
#include<set>
#include<vector>
#include<algorithm>
#include<string>
#include<queue>
using namespace std;
#define DEBUG(x) cout<<#x<<"="<<x<<endl
#define DEBUG2(x,y) cout<<#x<<"="<<x<<","<<#y<<"="<<y<<endl
typedef long long ll;
// #define LOCAL
const int MAXN=2e5+10;
int N;
vector<int>G[MAXN];
int c[MAXN];
ll ans[MAXN];
ll cal(int n)
{
    return (ll)n*(n+1)/2;
}
int num[MAXN];
int csz[MAXN];
void dfs(int rt,int fa)
{
    csz[rt]=1;
    int cl=c[rt];
    int t=num[cl];
    for(int v:G[rt]){
        if(v==fa)continue;
        int cur=num[cl];
        dfs(v,rt);
        int tmp=csz[v]-(num[cl]-cur);
        ans[cl]-=cal(tmp);
        csz[rt]+=csz[v];
    }
    num[cl]=t+csz[rt];
}
void solve()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin>>N;
    for(int u=1;u<=N;u++)cin>>c[u];
    for(int u=0;u<N-1;u++){
        int a,b;
        cin>>a>>b;
        G[a].push_back(b);
        G[b].push_back(a);
    }
    for(int c=1;c<=N;c++)ans[c]=cal(N);
    dfs(1,0);
    for(int c=1;c<=N;c++){
        ans[c]-=cal(N-num[c]);
        cout<<ans[c]<<'\n';
    }
}
int main()
{
#ifdef LOCAL
    freopen("in.txt","r",stdin);
#endif
    solve();
}
### Prompt

Generate 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<cstdio>
#include<cstring>
#include<cmath>
#include<cassert>
#include<iostream>
#include<iomanip>
#include<map>
#include<set>
#include<vector>
#include<algorithm>
#include<string>
#include<queue>
using namespace std;
#define DEBUG(x) cout<<#x<<"="<<x<<endl
#define DEBUG2(x,y) cout<<#x<<"="<<x<<","<<#y<<"="<<y<<endl
typedef long long ll;
// #define LOCAL
const int MAXN=2e5+10;
int N;
vector<int>G[MAXN];
int c[MAXN];
ll ans[MAXN];
ll cal(int n)
{
    return (ll)n*(n+1)/2;
}
int num[MAXN];
int csz[MAXN];
void dfs(int rt,int fa)
{
    csz[rt]=1;
    int cl=c[rt];
    int t=num[cl];
    for(int v:G[rt]){
        if(v==fa)continue;
        int cur=num[cl];
        dfs(v,rt);
        int tmp=csz[v]-(num[cl]-cur);
        ans[cl]-=cal(tmp);
        csz[rt]+=csz[v];
    }
    num[cl]=t+csz[rt];
}
void solve()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    cin>>N;
    for(int u=1;u<=N;u++)cin>>c[u];
    for(int u=0;u<N-1;u++){
        int a,b;
        cin>>a>>b;
        G[a].push_back(b);
        G[b].push_back(a);
    }
    for(int c=1;c<=N;c++)ans[c]=cal(N);
    dfs(1,0);
    for(int c=1;c<=N;c++){
        ans[c]-=cal(N-num[c]);
        cout<<ans[c]<<'\n';
    }
}
int main()
{
#ifdef LOCAL
    freopen("in.txt","r",stdin);
#endif
    solve();
}
```