#include <bits/stdc++.h>

#define int long long

using namespace std;

const int N = 2e5 + 5;
int occ[N], cnt[N], res[N], c[N], delt, n;
vector<int> g[N];

inline int square(int x) {
  return x * (x - 1) / 2;
}

void dfs(int u, int p = -1) {
  int before = cnt[c[u]] + delt;
  int parold = 0;
  if (p != -1) {
    parold = cnt[c[p]] + delt;
  }
  for (int v : g[u]) {
    if (v != p) {
      dfs(v, u);
    }
  }
  ++delt;
  cnt[c[u]] = -delt + before;
  if (p != -1) {
    res[c[p]] += square(cnt[c[p]] + delt - parold);
  } else {
    for (int i = 0; i < n; i++) {
      res[i] += square(cnt[i] + delt);
    }
  }
}

signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> c[i];
    --c[i];
    ++occ[c[i]];
  }
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    --u, --v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(0);
  for (int i = 0; i < n; i++) {
    cout << square(n) - res[i] + occ[i] << '\n';
  }
  return 0;
}

### Prompt

Your task is to create 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>

#define int long long

using namespace std;

const int N = 2e5 + 5;
int occ[N], cnt[N], res[N], c[N], delt, n;
vector<int> g[N];

inline int square(int x) {
  return x * (x - 1) / 2;
}

void dfs(int u, int p = -1) {
  int before = cnt[c[u]] + delt;
  int parold = 0;
  if (p != -1) {
    parold = cnt[c[p]] + delt;
  }
  for (int v : g[u]) {
    if (v != p) {
      dfs(v, u);
    }
  }
  ++delt;
  cnt[c[u]] = -delt + before;
  if (p != -1) {
    res[c[p]] += square(cnt[c[p]] + delt - parold);
  } else {
    for (int i = 0; i < n; i++) {
      res[i] += square(cnt[i] + delt);
    }
  }
}

signed main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) {
    cin >> c[i];
    --c[i];
    ++occ[c[i]];
  }
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    --u, --v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(0);
  for (int i = 0; i < n; i++) {
    cout << square(n) - res[i] + occ[i] << '\n';
  }
  return 0;
}

```