#include <bits/stdc++.h>
using namespace std;
 
const int maxn = 2e5 + 5;
 
int n;
int c[maxn];
vector<int> G[maxn];
 
int sz[maxn];
int tot;
int cnt[maxn];
typedef long long ll;
 
ll ans[maxn];
 
void dfs(int u, int fa){
    sz[u] = 1;
    int in = cnt[c[u]]; 
    for(auto v : G[u]){
        if(v == fa)
            continue;
        int front = tot - cnt[c[u]];
        dfs(v, u);
        sz[u] += sz[v];
        int back = tot - cnt[c[u]];
        ans[c[u]] -= 1LL * (back - front) * (back - front + 1) / 2;
    } 
    tot++;
    cnt[c[u]] = in + sz[u];
}
 
int main(){
    ios::sync_with_stdio(0), cin.tie(0), cout.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, 1);
    for(int i = 1;i <= n;i++){
        ans[i] -= 1LL * (n - cnt[i]) * (n - cnt[i] + 1) / 2;
        ans[i] += 1LL * n * (n + 1) / 2;
    }
    for(int i = 1;i <= n;i++){
        cout << ans[i] << endl;
    }
    return 0;
}
### 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 <bits/stdc++.h>
using namespace std;
 
const int maxn = 2e5 + 5;
 
int n;
int c[maxn];
vector<int> G[maxn];
 
int sz[maxn];
int tot;
int cnt[maxn];
typedef long long ll;
 
ll ans[maxn];
 
void dfs(int u, int fa){
    sz[u] = 1;
    int in = cnt[c[u]]; 
    for(auto v : G[u]){
        if(v == fa)
            continue;
        int front = tot - cnt[c[u]];
        dfs(v, u);
        sz[u] += sz[v];
        int back = tot - cnt[c[u]];
        ans[c[u]] -= 1LL * (back - front) * (back - front + 1) / 2;
    } 
    tot++;
    cnt[c[u]] = in + sz[u];
}
 
int main(){
    ios::sync_with_stdio(0), cin.tie(0), cout.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, 1);
    for(int i = 1;i <= n;i++){
        ans[i] -= 1LL * (n - cnt[i]) * (n - cnt[i] + 1) / 2;
        ans[i] += 1LL * n * (n + 1) / 2;
    }
    for(int i = 1;i <= n;i++){
        cout << ans[i] << endl;
    }
    return 0;
}
```