#include <bits/stdc++.h>
#define int long long
#define fi first
#define se second

using namespace std;

const int maxN = 2e5 + 10;

int n;
vector<int> adj[maxN];
int c[maxN], color[maxN], res[maxN], son[maxN];

void dfs(int u, int par)
{
    int cnt = c[color[u]];

    for(int v : adj[u])
    {
        if(v == par) continue;
        c[color[u]] = 0;
        dfs(v, u);
        son[u] += son[v];
        res[color[u]] += (son[v] - c[color[u]]) * (son[v] - c[color[u]] + 1) / 2;
    }
    c[color[u]] = cnt + son[u];
}

int32_t main()
{
    ios_base::sync_with_stdio(false); cin.tie(); cout.tie();

    cin >> n;
    for(int i = 1; i <= n; ++i){
        cin >> color[i];
        son[i] = 1;
    }
    for(int i = 1; i <= n - 1; ++i)
    {
        int u,v;
        cin >> u >> v;
        adj[u].push_back(v);
        adj[v].push_back(u);
    }

    dfs(1, -1);

    for(int i = 1; i <= n; ++i){
        cout << n * (n + 1) / 2 - (n - c[i]) * (n - c[i] + 1) / 2 - res[i] << endl;
    }

}

### Prompt

Develop a solution in Cpp 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>
#define int long long
#define fi first
#define se second

using namespace std;

const int maxN = 2e5 + 10;

int n;
vector<int> adj[maxN];
int c[maxN], color[maxN], res[maxN], son[maxN];

void dfs(int u, int par)
{
    int cnt = c[color[u]];

    for(int v : adj[u])
    {
        if(v == par) continue;
        c[color[u]] = 0;
        dfs(v, u);
        son[u] += son[v];
        res[color[u]] += (son[v] - c[color[u]]) * (son[v] - c[color[u]] + 1) / 2;
    }
    c[color[u]] = cnt + son[u];
}

int32_t main()
{
    ios_base::sync_with_stdio(false); cin.tie(); cout.tie();

    cin >> n;
    for(int i = 1; i <= n; ++i){
        cin >> color[i];
        son[i] = 1;
    }
    for(int i = 1; i <= n - 1; ++i)
    {
        int u,v;
        cin >> u >> v;
        adj[u].push_back(v);
        adj[v].push_back(u);
    }

    dfs(1, -1);

    for(int i = 1; i <= n; ++i){
        cout << n * (n + 1) / 2 - (n - c[i]) * (n - c[i] + 1) / 2 - res[i] << endl;
    }

}

```