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

long long n,a,b,s=0;
vector<long long> c, d, ans;
vector<vector<long long>> G;

long long ret(long long a){
    return a*(a+1)/2;
}

void dfs(long long u, long long prev) {
    long long t = s + d[c[u]];
    for(auto v: G[u]) {
        if (v == prev) {
            continue;
        }
        // cout<<"v: " << v<<endl;
        d[c[u]] = -s;
        dfs(v, u);
        ans[c[u]] -= ret(s + d[c[u]]);
    }
    s++;
    d[c[u]] = t-s;
}

int main() {
    cin>>n;
    
    G.assign(n, vector<long long>());
    c.assign(n, 0);
    d.assign(n, 0);
    ans.assign(n, ret(n));

    for(int i=0;i<n;i++) {
        long long cc;
        cin >>cc;
        c[i] = cc-1;
    }
    for(int i=0;i<n - 1;i++) {
        cin >>a>>b;
        a--;b--;
        G[a].push_back(b);
        G[b].push_back(a);
    }

    dfs(0, -1);
    for(int k=0;k<n;k++) {
        ans[k] = ans[k] - ret(s+d[k]);
    }
    for(int i=0;i<n;i++){
        cout<<ans[i]<<"\n";
    }
    return 0;
}
### Prompt

Construct a CPP code solution to the problem outlined:

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;

long long n,a,b,s=0;
vector<long long> c, d, ans;
vector<vector<long long>> G;

long long ret(long long a){
    return a*(a+1)/2;
}

void dfs(long long u, long long prev) {
    long long t = s + d[c[u]];
    for(auto v: G[u]) {
        if (v == prev) {
            continue;
        }
        // cout<<"v: " << v<<endl;
        d[c[u]] = -s;
        dfs(v, u);
        ans[c[u]] -= ret(s + d[c[u]]);
    }
    s++;
    d[c[u]] = t-s;
}

int main() {
    cin>>n;
    
    G.assign(n, vector<long long>());
    c.assign(n, 0);
    d.assign(n, 0);
    ans.assign(n, ret(n));

    for(int i=0;i<n;i++) {
        long long cc;
        cin >>cc;
        c[i] = cc-1;
    }
    for(int i=0;i<n - 1;i++) {
        cin >>a>>b;
        a--;b--;
        G[a].push_back(b);
        G[b].push_back(a);
    }

    dfs(0, -1);
    for(int k=0;k<n;k++) {
        ans[k] = ans[k] - ret(s+d[k]);
    }
    for(int i=0;i<n;i++){
        cout<<ans[i]<<"\n";
    }
    return 0;
}
```