#include <bits/stdc++.h>
using namespace std;
using ll = long long;
#ifdef tabr
#include "library/debug.cpp"
#else
#define debug(...)
#endif

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n;
    cin >> n;
    vector<int> c(n), d(n);
    for (int i = 0; i < n; i++) {
        cin >> c[i];
        c[i]--;
        d[c[i]]++;
    }
    vector<vector<int>> g(n);
    for (int i = 0; i < n - 1; i++) {
        int a, b;
        cin >> a >> b;
        a--, b--;
        g[a].emplace_back(b);
        g[b].emplace_back(a);
    }
    vector<ll> sz(n, 1), s(n), t(n), ans(n, (ll)n * (n + 1) / 2);
    function<void(int, int)> dfs = [&](int v, int p) {
        t[v] = s[c[v]];
        for (int to : g[v]) {
            if (to == p) continue;
            ll dif = -s[c[v]];
            dfs(to, v);
            dif += s[c[v]];
            sz[v] += sz[to];
            ll sum = sz[to] - dif;
            ans[c[v]] -= sum * (sum + 1) / 2;
        }
        s[c[v]] = t[v] + sz[v];
    };
    dfs(0, -1);
    for (int i = 0; i < n; i++) {
        s[i] = n - s[i];
        ans[i] -= s[i] * (s[i] + 1) / 2;
        cout << ans[i] << '\n';
    }
    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;
#ifdef tabr
#include "library/debug.cpp"
#else
#define debug(...)
#endif

int main() {
    ios::sync_with_stdio(false);
    cin.tie(0);
    int n;
    cin >> n;
    vector<int> c(n), d(n);
    for (int i = 0; i < n; i++) {
        cin >> c[i];
        c[i]--;
        d[c[i]]++;
    }
    vector<vector<int>> g(n);
    for (int i = 0; i < n - 1; i++) {
        int a, b;
        cin >> a >> b;
        a--, b--;
        g[a].emplace_back(b);
        g[b].emplace_back(a);
    }
    vector<ll> sz(n, 1), s(n), t(n), ans(n, (ll)n * (n + 1) / 2);
    function<void(int, int)> dfs = [&](int v, int p) {
        t[v] = s[c[v]];
        for (int to : g[v]) {
            if (to == p) continue;
            ll dif = -s[c[v]];
            dfs(to, v);
            dif += s[c[v]];
            sz[v] += sz[to];
            ll sum = sz[to] - dif;
            ans[c[v]] -= sum * (sum + 1) / 2;
        }
        s[c[v]] = t[v] + sz[v];
    };
    dfs(0, -1);
    for (int i = 0; i < n; i++) {
        s[i] = n - s[i];
        ans[i] -= s[i] * (s[i] + 1) / 2;
        cout << ans[i] << '\n';
    }
    return 0;
}
```