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

int64_t n;
int64_t c[200000];
vector<int> edges[200000];
int64_t cnt[200000], offset;
int64_t block[200000];

void dfs(int v, int p=-1){
    int memo = cnt[c[v]] + offset;
    ++offset;
    for(auto to: edges[v]){
        if(to == p) continue;
        cnt[c[v]] = -offset;
        dfs(to, v);
        int64_t b = cnt[c[v]]+offset;
        block[c[v]] += b * (b + 1) / 2;
    }
    cnt[c[v]] = memo - offset;
}

int main(){
    cin.tie(nullptr);
    ios::sync_with_stdio(false);

    cin >> n;
    for(int i=0;i<n;++i){ cin >> c[i]; --c[i]; }
    for(int i=1;i<n;++i){
        int a, b;
        cin >> a >> b;
        edges[a-1].push_back(b-1);
        edges[b-1].push_back(a-1);
    }
    dfs(0);
    for(int i=0;i<n;++i){
        int64_t b = cnt[i] + offset;
        block[i] += b * (b+1) / 2;
    }

    for(int i=0;i<n;++i){
        cout << n * (n+1) / 2 - block[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;

int64_t n;
int64_t c[200000];
vector<int> edges[200000];
int64_t cnt[200000], offset;
int64_t block[200000];

void dfs(int v, int p=-1){
    int memo = cnt[c[v]] + offset;
    ++offset;
    for(auto to: edges[v]){
        if(to == p) continue;
        cnt[c[v]] = -offset;
        dfs(to, v);
        int64_t b = cnt[c[v]]+offset;
        block[c[v]] += b * (b + 1) / 2;
    }
    cnt[c[v]] = memo - offset;
}

int main(){
    cin.tie(nullptr);
    ios::sync_with_stdio(false);

    cin >> n;
    for(int i=0;i<n;++i){ cin >> c[i]; --c[i]; }
    for(int i=1;i<n;++i){
        int a, b;
        cin >> a >> b;
        edges[a-1].push_back(b-1);
        edges[b-1].push_back(a-1);
    }
    dfs(0);
    for(int i=0;i<n;++i){
        int64_t b = cnt[i] + offset;
        block[i] += b * (b+1) / 2;
    }

    for(int i=0;i<n;++i){
        cout << n * (n+1) / 2 - block[i] << endl;
    }
    return 0;
}
```