#include <bits/stdc++.h>

using namespace std;
using ll=long long;

void solve() {
    int n; cin >> n;
    vector<int> a(n);
    for (auto& x: a) {
        cin >> x; --x;
    }
    vector<vector<int>> g(n);
    for (int _ = 1; _ < n; _++) {
        int x,y;
        cin >> x >> y;
        x--;y--;
        g[x].push_back(y);
        g[y].push_back(x);
    }

    vector<ll> res(n);
    vector<int> sz(n,1), cnt(n);
    auto calc = [&](ll x){
        return x*(x+1)/2;
    };
    function<void(int,int)> dfs = [&](int u, int p){
        int c = a[u];
        int sub = 0;
        for (int v: g[u])if(v!=p){
            int x = cnt[c];
            dfs(v, u);
            sz[u] += sz[v];
            x = cnt[c]-x;
            sub += x;
            res[c] += calc(sz[v] - x);
        }
        cnt[c] -= sub;
        cnt[c] += sz[u];
    };
    dfs(0,-1);
    for (int c = 0; c < n; c++) {
        res[c] += calc(n - cnt[c]);
    }
    const ll tot = calc(n);
    for (auto& x: res) {
        cout << tot-x << '\n';
    }
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    solve();
    return 0;
}

### Prompt

Please formulate 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;

void solve() {
    int n; cin >> n;
    vector<int> a(n);
    for (auto& x: a) {
        cin >> x; --x;
    }
    vector<vector<int>> g(n);
    for (int _ = 1; _ < n; _++) {
        int x,y;
        cin >> x >> y;
        x--;y--;
        g[x].push_back(y);
        g[y].push_back(x);
    }

    vector<ll> res(n);
    vector<int> sz(n,1), cnt(n);
    auto calc = [&](ll x){
        return x*(x+1)/2;
    };
    function<void(int,int)> dfs = [&](int u, int p){
        int c = a[u];
        int sub = 0;
        for (int v: g[u])if(v!=p){
            int x = cnt[c];
            dfs(v, u);
            sz[u] += sz[v];
            x = cnt[c]-x;
            sub += x;
            res[c] += calc(sz[v] - x);
        }
        cnt[c] -= sub;
        cnt[c] += sz[u];
    };
    dfs(0,-1);
    for (int c = 0; c < n; c++) {
        res[c] += calc(n - cnt[c]);
    }
    const ll tot = calc(n);
    for (auto& x: res) {
        cout << tot-x << '\n';
    }
}

int main() {
    ios_base::sync_with_stdio(false);
    cin.tie(NULL);
    solve();
    return 0;
}

```