#include <bits/stdc++.h>

using namespace std;

using ll = long long;
using Graph = vector< vector<int> >;

ll dfs(int cur, int par, const Graph& g, const vector<int>& c, vector<ll>& pena, vector<ll>& ans) {
    ll cnt = 0;
    ll prv_pena = pena[c[cur]];
    for (int nex : g[cur]) {
        if (nex == par) continue;
        pena[c[cur]] = 0;
        ll nex_cnt = dfs(nex, cur, g, c, pena, ans);
        ll tmp = nex_cnt - pena[c[cur]];
        ans[c[cur]] -= tmp * (tmp + 1) / 2;
        cnt += nex_cnt;
    }

    ++cnt;
    pena[c[cur]] = prv_pena + cnt;
    return cnt;
}

int main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    int n;
    cin >> n;

    vector<int> c(n);
    for (int i = 0; i < n; ++i) {
        cin >> c[i];
        --c[i];
    }

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

    vector<ll> pena(n, 0);
    vector<ll> ans(n, (ll)n * (n + 1) / 2);
    ll cnt = dfs(0, -1, g, c, pena, ans);

    for (int i = 0; i < n; ++i) {
        ll tmp = cnt - pena[i];
        ans[i] -= tmp * (tmp + 1) / 2;
        cout << ans[i] << "\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;

using ll = long long;
using Graph = vector< vector<int> >;

ll dfs(int cur, int par, const Graph& g, const vector<int>& c, vector<ll>& pena, vector<ll>& ans) {
    ll cnt = 0;
    ll prv_pena = pena[c[cur]];
    for (int nex : g[cur]) {
        if (nex == par) continue;
        pena[c[cur]] = 0;
        ll nex_cnt = dfs(nex, cur, g, c, pena, ans);
        ll tmp = nex_cnt - pena[c[cur]];
        ans[c[cur]] -= tmp * (tmp + 1) / 2;
        cnt += nex_cnt;
    }

    ++cnt;
    pena[c[cur]] = prv_pena + cnt;
    return cnt;
}

int main() {
    cin.tie(nullptr);
    ios::sync_with_stdio(false);
    int n;
    cin >> n;

    vector<int> c(n);
    for (int i = 0; i < n; ++i) {
        cin >> c[i];
        --c[i];
    }

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

    vector<ll> pena(n, 0);
    vector<ll> ans(n, (ll)n * (n + 1) / 2);
    ll cnt = dfs(0, -1, g, c, pena, ans);

    for (int i = 0; i < n; ++i) {
        ll tmp = cnt - pena[i];
        ans[i] -= tmp * (tmp + 1) / 2;
        cout << ans[i] << "\n";
    }
    return 0;
}
```