#include <bits/stdc++.h>
#define int long long
#define f first
#define s second
using namespace std;
const int N=2e5+9;
int n,c[N],ans[N],sz[N];
map <int,int> mp[N];
vector <int> g[N];
void dfs(int v,int p){
    int hv=-1,mx=0;
    for(auto u:g[v]){
        if(u==p)continue;
        dfs(u,v);
        sz[v]+=sz[u];
        int x=sz[u]-mp[u][c[v]];
        ans[c[v]]+=x*(x-1)/2+x;
        if(mp[u].size()>mx)mx=mp[u].size(),hv=u;
    }
    swap(mp[hv],mp[v]);
    for(auto u:g[v]){
        if(u==p or u==hv)continue;
        for(auto i:mp[u])
            mp[v][i.f]+=i.s;
    }
    sz[v]++;
    mp[v][c[v]]=sz[v];
}
int32_t main(){
    ios::sync_with_stdio(0),cin.tie(0);
    cin >> n;
    for(int i=1;i<=n;i++)
        cin >> c[i];
    for(int i=1;i<n;i++){
        int u,v;
        cin >> u >> v;
        g[u].push_back(v);
        g[v].push_back(u);
    }
    dfs(1,0);
    for(int i=1;i<=n;i++){
        int x=n-mp[1][i];
        ans[i]+=x*(x-1)/2+x;
    }
    int k=n*(n-1)/2+n;
    for(int i=1;i<=n;i++)
        cout << k-ans[i] << '\n';
}
### Prompt

Develop a solution in CPP to the problem described below:

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>
#define int long long
#define f first
#define s second
using namespace std;
const int N=2e5+9;
int n,c[N],ans[N],sz[N];
map <int,int> mp[N];
vector <int> g[N];
void dfs(int v,int p){
    int hv=-1,mx=0;
    for(auto u:g[v]){
        if(u==p)continue;
        dfs(u,v);
        sz[v]+=sz[u];
        int x=sz[u]-mp[u][c[v]];
        ans[c[v]]+=x*(x-1)/2+x;
        if(mp[u].size()>mx)mx=mp[u].size(),hv=u;
    }
    swap(mp[hv],mp[v]);
    for(auto u:g[v]){
        if(u==p or u==hv)continue;
        for(auto i:mp[u])
            mp[v][i.f]+=i.s;
    }
    sz[v]++;
    mp[v][c[v]]=sz[v];
}
int32_t main(){
    ios::sync_with_stdio(0),cin.tie(0);
    cin >> n;
    for(int i=1;i<=n;i++)
        cin >> c[i];
    for(int i=1;i<n;i++){
        int u,v;
        cin >> u >> v;
        g[u].push_back(v);
        g[v].push_back(u);
    }
    dfs(1,0);
    for(int i=1;i<=n;i++){
        int x=n-mp[1][i];
        ans[i]+=x*(x-1)/2+x;
    }
    int k=n*(n-1)/2+n;
    for(int i=1;i<=n;i++)
        cout << k-ans[i] << '\n';
}
```