#include <bits/stdc++.h>

using namespace std;

mt19937 rng(chrono::high_resolution_clock::now().time_since_epoch().count());
typedef long long ll;
typedef long double ld;

const int N = (int) 2e5 + 7;
int n, col[N], sub[N], pr[N];
vector<int> g[N];
ll su[N], sol[N];

void build(int a) {
  sub[a] = 1;
  vector<int> kids;
  for (auto &b : g[a]) {
    if (sub[b] == 0) {
      pr[b] = a;
      kids.push_back(b);
      build(b);
      sub[a] += sub[b];
    }
  }
  g[a] = kids;
}

ll f(ll len) {
  return len * (len + 1) / 2;
}

void dfs(int a) {
  int ini1 = su[col[a]];
  int ini2 = su[col[pr[a]]];
  for (auto &b : g[a]) {
    dfs(b);
  }
  su[col[a]] = ini1 + sub[a];
  int tot = su[col[pr[a]]] - ini2;
  tot = sub[a] - tot;
  sol[col[pr[a]]] += f(tot);
}

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> col[i];
  }
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  build(1);
  dfs(1);
  for (int i = 1; i <= n; i++) {
    sol[i] += f(sub[1] - su[i]);
    sol[i] = f(n) - sol[i];
    cout << sol[i] << "\n";
  }
  return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;

mt19937 rng(chrono::high_resolution_clock::now().time_since_epoch().count());
typedef long long ll;
typedef long double ld;

const int N = (int) 2e5 + 7;
int n, col[N], sub[N], pr[N];
vector<int> g[N];
ll su[N], sol[N];

void build(int a) {
  sub[a] = 1;
  vector<int> kids;
  for (auto &b : g[a]) {
    if (sub[b] == 0) {
      pr[b] = a;
      kids.push_back(b);
      build(b);
      sub[a] += sub[b];
    }
  }
  g[a] = kids;
}

ll f(ll len) {
  return len * (len + 1) / 2;
}

void dfs(int a) {
  int ini1 = su[col[a]];
  int ini2 = su[col[pr[a]]];
  for (auto &b : g[a]) {
    dfs(b);
  }
  su[col[a]] = ini1 + sub[a];
  int tot = su[col[pr[a]]] - ini2;
  tot = sub[a] - tot;
  sol[col[pr[a]]] += f(tot);
}

int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> col[i];
  }
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  build(1);
  dfs(1);
  for (int i = 1; i <= n; i++) {
    sol[i] += f(sub[1] - su[i]);
    sol[i] = f(n) - sol[i];
    cout << sol[i] << "\n";
  }
  return 0;
}

```