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

typedef long long ll;

ll n;
vector<int> g[200005];
int c[200005];
int sz[200005];
ll cnt[200005],tot;
ll ans[200005];

int dfs(int x,int p){
    sz[x]=1;
    int &col=c[x];
    ll tmp=cnt[col];
    for(int &y:g[x])if(y!=p){
        ll pre=tot-cnt[col];
        sz[x]+=dfs(y,x);
        ll cur=tot-cnt[col]-pre;
        ans[col]+=cur*(cur-1)/2;
    }
    cnt[col]=tmp+sz[x];
    tot++;
    return sz[x];
}

int main(){

    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    cin>>n;
    for(int i=1;i<=n;i++){
        cin>>c[i];ans[c[i]]--;
    }
    for(int i=1;i<n;i++){
        int a,b;
        cin>>a>>b;
        g[a].push_back(b);
        g[b].push_back(a);
    }
    dfs(1,-1);
    for(int i=1;i<=n;i++){
        ans[i]+=(n-cnt[i])*(n-cnt[i]-1)/2;
        cout<<n*(n-1)/2-ans[i]<<'\n';
    }

    return 0;
}
### 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<bits/stdc++.h>
using namespace std;

typedef long long ll;

ll n;
vector<int> g[200005];
int c[200005];
int sz[200005];
ll cnt[200005],tot;
ll ans[200005];

int dfs(int x,int p){
    sz[x]=1;
    int &col=c[x];
    ll tmp=cnt[col];
    for(int &y:g[x])if(y!=p){
        ll pre=tot-cnt[col];
        sz[x]+=dfs(y,x);
        ll cur=tot-cnt[col]-pre;
        ans[col]+=cur*(cur-1)/2;
    }
    cnt[col]=tmp+sz[x];
    tot++;
    return sz[x];
}

int main(){

    ios::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);

    cin>>n;
    for(int i=1;i<=n;i++){
        cin>>c[i];ans[c[i]]--;
    }
    for(int i=1;i<n;i++){
        int a,b;
        cin>>a>>b;
        g[a].push_back(b);
        g[b].push_back(a);
    }
    dfs(1,-1);
    for(int i=1;i<=n;i++){
        ans[i]+=(n-cnt[i])*(n-cnt[i]-1)/2;
        cout<<n*(n-1)/2-ans[i]<<'\n';
    }

    return 0;
}
```