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

typedef long long LL;

const int N = 200003;
int n, c[N];
vector<int> graph[N];

int clk, d[N], proc[N];

LL dp[N];

void dfs(int cur, int par = 0) {
    int col = c[cur];
    d[cur] = 1;
    for(int child : graph[cur]) {
        if(child == par) continue;

        LL before = proc[col];
        dfs(child, cur);
        d[cur] += d[child];
        LL after = proc[col];
        LL comp = d[child] - (after-before);
        cerr << col << ' ' << comp << endl;
        
        dp[col] += comp*(comp-1)/2 + comp;
        proc[col] += comp;
    }
    proc[col]++;
}

int main() {

    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

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

    dfs(1);

    LL tot = (LL)n*(n-1)/2 + n;
    for(int i = 1; i <= n; ++i) {
        LL rem = n - proc[i];
        dp[i] += rem*(rem-1)/2 + rem;
        LL ans = tot - dp[i];
        cout << ans << '\n';
    }

    return 0;
}

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

typedef long long LL;

const int N = 200003;
int n, c[N];
vector<int> graph[N];

int clk, d[N], proc[N];

LL dp[N];

void dfs(int cur, int par = 0) {
    int col = c[cur];
    d[cur] = 1;
    for(int child : graph[cur]) {
        if(child == par) continue;

        LL before = proc[col];
        dfs(child, cur);
        d[cur] += d[child];
        LL after = proc[col];
        LL comp = d[child] - (after-before);
        cerr << col << ' ' << comp << endl;
        
        dp[col] += comp*(comp-1)/2 + comp;
        proc[col] += comp;
    }
    proc[col]++;
}

int main() {

    ios_base::sync_with_stdio(false);
    cin.tie(NULL);

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

    dfs(1);

    LL tot = (LL)n*(n-1)/2 + n;
    for(int i = 1; i <= n; ++i) {
        LL rem = n - proc[i];
        dp[i] += rem*(rem-1)/2 + rem;
        LL ans = tot - dp[i];
        cout << ans << '\n';
    }

    return 0;
}

```