#include <bits/stdc++.h>
#include <iostream>
using namespace std;
#define _LL long long
#define rep(i, n) for (_LL i = 0; i < (_LL)(n); i++)

#define NMAX 200000
vector<int> b[NMAX];
_LL sum[NMAX];
_LL c[NMAX];
_LL cc[NMAX];
_LL totalcount = 0;
 
void search(int current, int p)
{
    _LL index = 1;
    _LL pcc = cc[c[current]] + totalcount;
    totalcount++;

    for( auto itr = b[current].begin(); itr != b[current].end(); itr++ )
    {
        if( *itr != p ) 
        {
            cc[c[current]] = -totalcount;
            search(*itr, current);
            _LL subc = cc[c[current]] + totalcount;
            sum[c[current]] += subc * (subc + 1) / 2;
        }
    }

    cc[c[current]] = pcc - totalcount;
}
 
int main()
{
    _LL n;
    cin >> n;
    rep(i, n) 
    {
        cin >> c[i];
        c[i]--;
    }
    rep(i, n) sum[i] = 0;
    rep(i, n) cc[i] = 0;
    rep(i, n - 1)
    {
        int u, v;
        cin >> u >> v;
        u--; v--;
        b[u].push_back(v);
        b[v].push_back(u);
    }
    search(0, -1);
    rep(i, n)
    {
            _LL subc = cc[i] + totalcount;
            sum[i] += subc * (subc + 1) / 2;  
    }

    _LL total = n * (n + 1) / 2;

    rep(i, n) cout << total - sum[i] << endl;
    
}
### Prompt

Your challenge is to write 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>
#include <iostream>
using namespace std;
#define _LL long long
#define rep(i, n) for (_LL i = 0; i < (_LL)(n); i++)

#define NMAX 200000
vector<int> b[NMAX];
_LL sum[NMAX];
_LL c[NMAX];
_LL cc[NMAX];
_LL totalcount = 0;
 
void search(int current, int p)
{
    _LL index = 1;
    _LL pcc = cc[c[current]] + totalcount;
    totalcount++;

    for( auto itr = b[current].begin(); itr != b[current].end(); itr++ )
    {
        if( *itr != p ) 
        {
            cc[c[current]] = -totalcount;
            search(*itr, current);
            _LL subc = cc[c[current]] + totalcount;
            sum[c[current]] += subc * (subc + 1) / 2;
        }
    }

    cc[c[current]] = pcc - totalcount;
}
 
int main()
{
    _LL n;
    cin >> n;
    rep(i, n) 
    {
        cin >> c[i];
        c[i]--;
    }
    rep(i, n) sum[i] = 0;
    rep(i, n) cc[i] = 0;
    rep(i, n - 1)
    {
        int u, v;
        cin >> u >> v;
        u--; v--;
        b[u].push_back(v);
        b[v].push_back(u);
    }
    search(0, -1);
    rep(i, n)
    {
            _LL subc = cc[i] + totalcount;
            sum[i] += subc * (subc + 1) / 2;  
    }

    _LL total = n * (n + 1) / 2;

    rep(i, n) cout << total - sum[i] << endl;
    
}
```