#include <bits/stdc++.h>
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  int n;
  std::cin >> n;
  std::vector<int> a(n), sz(n), tag(n); 
  std::vector<std::vector<int>> g(n);
  std::vector<long long> ans(n);
  std::function<long long(int)> calc = [&](int x) {
    return 1ll * x * (x + 1) / 2;
  };
  for (int i = 0; i < n; ++i) {
    std::cin >> a[i];
    --a[i];
  }
  for (int i = 0; i < n; ++i)
    ans[i] = calc(n);  
  for (int i = 1; i < n; ++i) {
    int u, v;
    std::cin >> u >> v;
    --u;
    --v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  std::function<void(int, int)> dfs = [&](int u, int fa) {
    sz[u] = 1;
    int lst = tag[a[u]];
    for (auto v : g[u]) {
      if (v == fa) 
        continue;
      int pre = tag[a[u]];
      dfs(v, u);
      sz[u] += sz[v];
      ans[a[u]] -= calc(sz[v] - (tag[a[u]] - pre));
    }
    tag[a[u]] = sz[u] + lst;
  };   
  dfs(0, -1);
  for (int i = 0; i < n; ++i)
    ans[i] -= calc(n - tag[i]);

  for (int i = 0; i < n; ++i)
    std::cout << ans[i] << "\n";
  return 0;
}
### Prompt

In cpp, your task is to solve 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>
int main() {
  std::ios::sync_with_stdio(false);
  std::cin.tie(nullptr);
  int n;
  std::cin >> n;
  std::vector<int> a(n), sz(n), tag(n); 
  std::vector<std::vector<int>> g(n);
  std::vector<long long> ans(n);
  std::function<long long(int)> calc = [&](int x) {
    return 1ll * x * (x + 1) / 2;
  };
  for (int i = 0; i < n; ++i) {
    std::cin >> a[i];
    --a[i];
  }
  for (int i = 0; i < n; ++i)
    ans[i] = calc(n);  
  for (int i = 1; i < n; ++i) {
    int u, v;
    std::cin >> u >> v;
    --u;
    --v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  std::function<void(int, int)> dfs = [&](int u, int fa) {
    sz[u] = 1;
    int lst = tag[a[u]];
    for (auto v : g[u]) {
      if (v == fa) 
        continue;
      int pre = tag[a[u]];
      dfs(v, u);
      sz[u] += sz[v];
      ans[a[u]] -= calc(sz[v] - (tag[a[u]] - pre));
    }
    tag[a[u]] = sz[u] + lst;
  };   
  dfs(0, -1);
  for (int i = 0; i < n; ++i)
    ans[i] -= calc(n - tag[i]);

  for (int i = 0; i < n; ++i)
    std::cout << ans[i] << "\n";
  return 0;
}
```