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

const int SIZE = 200000;
using ll = long long;
ll N;
ll all = 0;
vector<ll> colors(SIZE), cnt(SIZE, 0), sub(SIZE, 0), ans(SIZE, 0);
vector<vector<ll>> T(SIZE);

ll f(ll x){
    return x * (x-1) / 2;
}

ll dfs(int node = 0, int p = -1){
    sub[node] = 1;
    ll bfr = cnt[colors[node]];
    for (ll i : T[node]){
        if (i == p) continue;
        ll pre = all - cnt[colors[node]];
        sub[node] += dfs(i, node);
        ll next = all - cnt[colors[node]];
        ans[colors[node]] -= f(next - pre);
    }
    cnt[colors[node]] = bfr + sub[node];
    all++;
    return sub[node];
}

int main(){
    cin >> N;
    for (int i = 0; i < N; i++){
        cin >> colors[i];
        colors[i]--;
        ans[i] += f(N);
    }

    for (int i = 0; i < N-1; i++){
        int a, b; cin >> a >> b;
        a--;b--;
        T[a].push_back(b);
        T[b].push_back(a);
    }
    dfs();
    for (int color = 0; color < N; color++)
        ans[color] -= f(N-cnt[color]);
    for (int i = 0; i < N; i++) ans[colors[i]]++;
    for (int i = 0; i < N; i++) cout << ans[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>
using namespace std;

const int SIZE = 200000;
using ll = long long;
ll N;
ll all = 0;
vector<ll> colors(SIZE), cnt(SIZE, 0), sub(SIZE, 0), ans(SIZE, 0);
vector<vector<ll>> T(SIZE);

ll f(ll x){
    return x * (x-1) / 2;
}

ll dfs(int node = 0, int p = -1){
    sub[node] = 1;
    ll bfr = cnt[colors[node]];
    for (ll i : T[node]){
        if (i == p) continue;
        ll pre = all - cnt[colors[node]];
        sub[node] += dfs(i, node);
        ll next = all - cnt[colors[node]];
        ans[colors[node]] -= f(next - pre);
    }
    cnt[colors[node]] = bfr + sub[node];
    all++;
    return sub[node];
}

int main(){
    cin >> N;
    for (int i = 0; i < N; i++){
        cin >> colors[i];
        colors[i]--;
        ans[i] += f(N);
    }

    for (int i = 0; i < N-1; i++){
        int a, b; cin >> a >> b;
        a--;b--;
        T[a].push_back(b);
        T[b].push_back(a);
    }
    dfs();
    for (int color = 0; color < N; color++)
        ans[color] -= f(N-cnt[color]);
    for (int i = 0; i < N; i++) ans[colors[i]]++;
    for (int i = 0; i < N; i++) cout << ans[i] << endl;
    
}
```