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

ll N;
int c[200010];
int cnt[200010];
int son[200010];
ll ans[200010];
vector<int> G[200010];

void dfs(int v, int p){
    son[v] = 1;
    for(auto u : G[v]){
        if(u == p) continue;
        ll memo = cnt[c[v]];
        dfs(u, v);
        ll m = son[u] - (cnt[c[v]] - memo);
        ans[c[v]] -= m * (m - 1) / 2 + m;
        son[v] += son[u];
        cnt[c[v]] = memo + son[u];
    }
    cnt[c[v]]++;
}

int main(){
    cin >> N;
    for(int i = 1; i <= N; i++){
        cin >> c[i];
        ans[i] = N * (N - 1) / 2 + N;
    }
    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, 0);
    for(int i = 1; i <= N; i++){
        ll m = son[1] - cnt[i];
        ans[i] -= m * (m - 1) / 2 + m;
    }
    for(int i = 1; i <= N; i++) cout << ans[i] << endl;
    return 0;
}
### Prompt

Please provide a Cpp coded solution 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>
using namespace std;
#define ll long long

ll N;
int c[200010];
int cnt[200010];
int son[200010];
ll ans[200010];
vector<int> G[200010];

void dfs(int v, int p){
    son[v] = 1;
    for(auto u : G[v]){
        if(u == p) continue;
        ll memo = cnt[c[v]];
        dfs(u, v);
        ll m = son[u] - (cnt[c[v]] - memo);
        ans[c[v]] -= m * (m - 1) / 2 + m;
        son[v] += son[u];
        cnt[c[v]] = memo + son[u];
    }
    cnt[c[v]]++;
}

int main(){
    cin >> N;
    for(int i = 1; i <= N; i++){
        cin >> c[i];
        ans[i] = N * (N - 1) / 2 + N;
    }
    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, 0);
    for(int i = 1; i <= N; i++){
        ll m = son[1] - cnt[i];
        ans[i] -= m * (m - 1) / 2 + m;
    }
    for(int i = 1; i <= N; i++) cout << ans[i] << endl;
    return 0;
}
```