#include <bits/stdc++.h>
using namespace std;
const int N = 2e5 + 5;
int col[N] = {};
long long cnt[N] = {};
long long ans[N] = {};
vector<int> G[N] = {};
long long n;
long long dfs(int u, int f) {
    long long ret = 1, s = 0, tag = cnt[col[u]];
    long long intag = tag;
    for (const int &v : G[u]) {
        if (v == f) continue;
        s = dfs(v, u);
        ret += s;
        long long d = s - (cnt[col[u]] - tag);
        ans[col[u]] -= d * (d - 1) / 2;
        tag = cnt[col[u]]; 
    }
    cnt[col[u]] = intag + ret;
    return ret;
}
int main() {
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    cin >> n;
    fill(ans + 1, ans + 1 + n, n * (n - 1) / 2);
    for (int i = 1; i <= n; i++) {
        cin >> col[i];
        ans[col[i]]++;
    }
    int u, v;
    for (int i = 1; i < n; i++) {
        cin >> u >> v;
        G[u].emplace_back(v);
        G[v].emplace_back(u);
    }
    dfs(1, 1);
    for (int i = 1; i <= n; i++) {
        long long s = n - cnt[i];
        ans[i] -= s * (s - 1) / 2;
        cout << ans[i] << '\n';
    }
    return 0;
}
### 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 N = 2e5 + 5;
int col[N] = {};
long long cnt[N] = {};
long long ans[N] = {};
vector<int> G[N] = {};
long long n;
long long dfs(int u, int f) {
    long long ret = 1, s = 0, tag = cnt[col[u]];
    long long intag = tag;
    for (const int &v : G[u]) {
        if (v == f) continue;
        s = dfs(v, u);
        ret += s;
        long long d = s - (cnt[col[u]] - tag);
        ans[col[u]] -= d * (d - 1) / 2;
        tag = cnt[col[u]]; 
    }
    cnt[col[u]] = intag + ret;
    return ret;
}
int main() {
    ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
    cin >> n;
    fill(ans + 1, ans + 1 + n, n * (n - 1) / 2);
    for (int i = 1; i <= n; i++) {
        cin >> col[i];
        ans[col[i]]++;
    }
    int u, v;
    for (int i = 1; i < n; i++) {
        cin >> u >> v;
        G[u].emplace_back(v);
        G[v].emplace_back(u);
    }
    dfs(1, 1);
    for (int i = 1; i <= n; i++) {
        long long s = n - cnt[i];
        ans[i] -= s * (s - 1) / 2;
        cout << ans[i] << '\n';
    }
    return 0;
}
```