#include <bits/stdc++.h>

using namespace std;

const int maxn = 2e5+10;

typedef long long ll;

ll n, c[maxn], sz[maxn], used[maxn], ans[maxn];

vector<int> grafo[maxn];

void dfs(int u, int p = 0){
    sz[u] = 1;
    for(auto v : grafo[u]){
        if(v == p) continue;
        dfs(v, u);
        sz[u] += sz[v];
    }
}

void solve(int u, int p = 0){
    ans[c[u]] += (n - sz[u] - used[c[u]] + 1) * sz[u];

    ll sum = sz[u] - 1;

    for(auto v : grafo[u]){
        if(v == p) continue;
        sum -= sz[v];
        ans[c[u]] += sz[v] * sum;
    }

    ll previous_used = used[c[u]];

    for(auto v : grafo[u]){
        if(v == p) continue;
        used[c[u]] = n - sz[v];
        solve(v, u);
    }

    used[c[u]] = previous_used + sz[u];
}

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

    cin >> n;

    for(int i=1; i<=n; i++) cin >> c[i];

    for(int i=1; i<n; i++){
        int u, v;

        cin >> u >> v;

        grafo[u].push_back(v);
        grafo[v].push_back(u);
    }

    dfs(1);
    solve(1);

    for(int i=1; i<=n; i++) cout << ans[i] << "\n";

    return 0;
}

### Prompt

Please create a solution in cpp 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;

const int maxn = 2e5+10;

typedef long long ll;

ll n, c[maxn], sz[maxn], used[maxn], ans[maxn];

vector<int> grafo[maxn];

void dfs(int u, int p = 0){
    sz[u] = 1;
    for(auto v : grafo[u]){
        if(v == p) continue;
        dfs(v, u);
        sz[u] += sz[v];
    }
}

void solve(int u, int p = 0){
    ans[c[u]] += (n - sz[u] - used[c[u]] + 1) * sz[u];

    ll sum = sz[u] - 1;

    for(auto v : grafo[u]){
        if(v == p) continue;
        sum -= sz[v];
        ans[c[u]] += sz[v] * sum;
    }

    ll previous_used = used[c[u]];

    for(auto v : grafo[u]){
        if(v == p) continue;
        used[c[u]] = n - sz[v];
        solve(v, u);
    }

    used[c[u]] = previous_used + sz[u];
}

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

    cin >> n;

    for(int i=1; i<=n; i++) cin >> c[i];

    for(int i=1; i<n; i++){
        int u, v;

        cin >> u >> v;

        grafo[u].push_back(v);
        grafo[v].push_back(u);
    }

    dfs(1);
    solve(1);

    for(int i=1; i<=n; i++) cout << ans[i] << "\n";

    return 0;
}

```