#include <iostream>
#include <vector>
#include <functional>
using namespace std;

int main()
{
    int N;
    cin>>N;
    vector<int> C(N);
    for (int &c: C)
        cin>>c, c--;
    vector<vector<int>> E(N);
    for (int i=0; i<N-1; i++)
    {
        int a, b;
        cin>>a>>b;
        E[a-1].push_back(b-1);
        E[b-1].push_back(a-1);
    }

    vector<long long> A(N, (long long)N*(N+1)/2);
    vector<int> CN(N);
    function<int(int, int)> f = [&](int c, int p)
    {
        int s = 0;
        int sc = 0;
        for (int e: E[c])
            if (e!=p)
            {
                int pre = CN[C[c]];
                int n = f(e, c);
                long long nc = n-(CN[C[c]]-pre);
                A[C[c]] -= nc*(nc+1)/2;
                sc += nc;
                s += n;
            }
        CN[C[c]] += sc+1;
        return s+1;
    };
    f(0, -1);
    for (int c=0; c<N; c++)
    {
        long long t = N-CN[c];
        A[c] -= t*(t+1)/2;
    }
    for (long long a: A)
        cout<<a<<endl;
}

### Prompt

Generate 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 <iostream>
#include <vector>
#include <functional>
using namespace std;

int main()
{
    int N;
    cin>>N;
    vector<int> C(N);
    for (int &c: C)
        cin>>c, c--;
    vector<vector<int>> E(N);
    for (int i=0; i<N-1; i++)
    {
        int a, b;
        cin>>a>>b;
        E[a-1].push_back(b-1);
        E[b-1].push_back(a-1);
    }

    vector<long long> A(N, (long long)N*(N+1)/2);
    vector<int> CN(N);
    function<int(int, int)> f = [&](int c, int p)
    {
        int s = 0;
        int sc = 0;
        for (int e: E[c])
            if (e!=p)
            {
                int pre = CN[C[c]];
                int n = f(e, c);
                long long nc = n-(CN[C[c]]-pre);
                A[C[c]] -= nc*(nc+1)/2;
                sc += nc;
                s += n;
            }
        CN[C[c]] += sc+1;
        return s+1;
    };
    f(0, -1);
    for (int c=0; c<N; c++)
    {
        long long t = N-CN[c];
        A[c] -= t*(t+1)/2;
    }
    for (long long a: A)
        cout<<a<<endl;
}

```