#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=2e5+5;
vector<ll>g[N];
ll n,u,v,color[N],c[N],son[N],res[N]={0};
ll f(ll n) {return n*(n+1)/2;}
void dfs(ll u,ll fa){
    auto tmp=c[color[u]];
    for(auto v:g[u]){
        if(v==fa) continue;
        c[color[u]]=0;
        dfs(v,u);
        son[u]+=son[v];
        res[color[u]]+=f(son[v]-c[color[u]]);
    }
    c[color[u]]=tmp+son[u];
}

int main(){
    cin>>n;
    for(ll i=1;i<=n;i++){
        son[i]=1;
        cin>>color[i];
    }
    for(ll i=1;i<n;i++){
        cin>>u>>v;
        g[u].push_back(v);
        g[v].push_back(u);
    }
    dfs(1,-1);
    for(int i=1;i<=n;i++){
        cout<<f(n)-f(n-c[i])-res[i]<<endl;
    }
    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;
const int N=2e5+5;
vector<ll>g[N];
ll n,u,v,color[N],c[N],son[N],res[N]={0};
ll f(ll n) {return n*(n+1)/2;}
void dfs(ll u,ll fa){
    auto tmp=c[color[u]];
    for(auto v:g[u]){
        if(v==fa) continue;
        c[color[u]]=0;
        dfs(v,u);
        son[u]+=son[v];
        res[color[u]]+=f(son[v]-c[color[u]]);
    }
    c[color[u]]=tmp+son[u];
}

int main(){
    cin>>n;
    for(ll i=1;i<=n;i++){
        son[i]=1;
        cin>>color[i];
    }
    for(ll i=1;i<n;i++){
        cin>>u>>v;
        g[u].push_back(v);
        g[v].push_back(u);
    }
    dfs(1,-1);
    for(int i=1;i<=n;i++){
        cout<<f(n)-f(n-c[i])-res[i]<<endl;
    }
    return 0;
}

```