#include <bits/stdc++.h>
using namespace std;

void solve(std::istream &in, std::ostream &out);

int main() {
#ifdef LOCAL
  freopen("../IO/f.in", "r", stdin);
//  freopen("../IO/f.out", "w", stdout);
#else
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(nullptr);
  std::cout.tie(nullptr);
#endif
  solve(std::cin, std::cout);
  return 0;
}

constexpr int N = (int)2e5 + 5;
vector<int> g[N];
int c[N], cc[N];
int64_t ans[N];
int n;

int64_t calc(int64_t x) {
  return x * (x + 1) / 2;
}

int dfs(int v, int p = -1) {
  int sz = 1;
  int col = c[v];
  int prev = cc[col];
  int p1 = prev;
  for (int u : g[v]) if (u != p) {
    int s_sz = dfs(u, v);
    int diff = cc[col] - p1;
    ans[col] += calc(s_sz - diff);
    p1 = cc[col];
    sz += s_sz;
  }
  cc[col] = prev + sz;
  return sz;
}

void solve(std::istream &in, std::ostream &out) {
  in >> n;
  for (int i = 0; i < n; ++i) in >> c[i];
  for (int i = 0, u, v; i < n - 1; ++i) {
    in >> u >> v;
    --u;
    --v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(0);
  for (int i = 1; i <= n; ++i) {
    if (c[0] != i) {
      out << calc(n) - ans[i] - calc(n - cc[i]) << '\n';
    } else {
      out << calc(n) - ans[i] << '\n';
    }
  }
}

### Prompt

Create a solution in cpp for 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;

void solve(std::istream &in, std::ostream &out);

int main() {
#ifdef LOCAL
  freopen("../IO/f.in", "r", stdin);
//  freopen("../IO/f.out", "w", stdout);
#else
  std::ios_base::sync_with_stdio(false);
  std::cin.tie(nullptr);
  std::cout.tie(nullptr);
#endif
  solve(std::cin, std::cout);
  return 0;
}

constexpr int N = (int)2e5 + 5;
vector<int> g[N];
int c[N], cc[N];
int64_t ans[N];
int n;

int64_t calc(int64_t x) {
  return x * (x + 1) / 2;
}

int dfs(int v, int p = -1) {
  int sz = 1;
  int col = c[v];
  int prev = cc[col];
  int p1 = prev;
  for (int u : g[v]) if (u != p) {
    int s_sz = dfs(u, v);
    int diff = cc[col] - p1;
    ans[col] += calc(s_sz - diff);
    p1 = cc[col];
    sz += s_sz;
  }
  cc[col] = prev + sz;
  return sz;
}

void solve(std::istream &in, std::ostream &out) {
  in >> n;
  for (int i = 0; i < n; ++i) in >> c[i];
  for (int i = 0, u, v; i < n - 1; ++i) {
    in >> u >> v;
    --u;
    --v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(0);
  for (int i = 1; i <= n; ++i) {
    if (c[0] != i) {
      out << calc(n) - ans[i] - calc(n - cc[i]) << '\n';
    } else {
      out << calc(n) - ans[i] << '\n';
    }
  }
}

```