#include <iostream>
#include <vector>
using namespace std;
typedef long long ll;
ll c[200010],sz[200010],ans[200010],ans2[200010],par[200010];
vector<int> G[200010],col[200010],v[200010];
int dfs(int s,int p){
    par[s] = p;
    int x = v[c[s]].back(),cnt = 1;
    col[x].push_back(s);
    for(int x:G[s]){
        if(x==p) continue;
        v[c[s]].push_back(x);
        cnt += dfs(x,s);
        v[c[s]].pop_back();
    }
    return sz[s] = cnt;
}

int main(){
    ll i,n;
    cin >> n;
    for(i=0;i<n;i++){
        cin >> c[i]; c[i]--;
    }
    for(i=0;i<n - 1;i++){
        int a,b; cin >> a >> b; a--; b--;
        G[a].push_back(b); G[b].push_back(a);
    }
    for(i=0;i<n;i++){
        v[i].push_back(0);
        ans2[i] = n;
    }
    dfs(0,-1);
    for(i=1;i<n;i++){
        ll num = sz[i];
        for(int x:col[i]){
            num -= sz[x];
        }
        ans[c[par[i]]] += num*(num + 1)/2;
    }
    for(int x:col[0]){
        int y = c[x];
        ans2[y] -= sz[x];
    }
    for(i=0;i<n;i++){
        cout << n*(n + 1)/2 - ans2[i]*(ans2[i] + 1)/2 - ans[i] << endl;
    }
}
### Prompt

Please formulate 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 <iostream>
#include <vector>
using namespace std;
typedef long long ll;
ll c[200010],sz[200010],ans[200010],ans2[200010],par[200010];
vector<int> G[200010],col[200010],v[200010];
int dfs(int s,int p){
    par[s] = p;
    int x = v[c[s]].back(),cnt = 1;
    col[x].push_back(s);
    for(int x:G[s]){
        if(x==p) continue;
        v[c[s]].push_back(x);
        cnt += dfs(x,s);
        v[c[s]].pop_back();
    }
    return sz[s] = cnt;
}

int main(){
    ll i,n;
    cin >> n;
    for(i=0;i<n;i++){
        cin >> c[i]; c[i]--;
    }
    for(i=0;i<n - 1;i++){
        int a,b; cin >> a >> b; a--; b--;
        G[a].push_back(b); G[b].push_back(a);
    }
    for(i=0;i<n;i++){
        v[i].push_back(0);
        ans2[i] = n;
    }
    dfs(0,-1);
    for(i=1;i<n;i++){
        ll num = sz[i];
        for(int x:col[i]){
            num -= sz[x];
        }
        ans[c[par[i]]] += num*(num + 1)/2;
    }
    for(int x:col[0]){
        int y = c[x];
        ans2[y] -= sz[x];
    }
    for(i=0;i<n;i++){
        cout << n*(n + 1)/2 - ans2[i]*(ans2[i] + 1)/2 - ans[i] << endl;
    }
}
```