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

const int N = 200005;

int n;
int c[N];
vector<int> edges[N];

long long ans[N];

int sz[N];
map<int, int> colors[N];
void dfs(int u, int p) {
  sz[u] = 1;
  int heavy = u;
  for (int v: edges[u]) {
    if (v == p) continue;
    dfs(v, u);

    sz[u] += sz[v];
    if (heavy == u || colors[heavy].size() < colors[v].size()) {
      heavy = v;
    }

    int comp = sz[v] - colors[v][c[u]];
    ans[c[u]] -= 1LL * comp * (comp + 1) / 2;
    colors[v].erase(c[u]);
  }

  swap(colors[u], colors[heavy]);

  for (int v: edges[u]) {
    if (v == p) continue;

    for (pair<int, int> p: colors[v]) {
      colors[u][p.first] += p.second;
    }
  }

  colors[u][c[u]] += sz[u];
}

int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &c[i]);
    --c[i];
    ans[i] = 1LL * n * (n+1) / 2;
  }
  for (int i = 1; i < n; ++i) {
    int u, v;
    scanf("%d %d", &u, &v);
    --u, --v;

    edges[u].push_back(v);
    edges[v].push_back(u);
  }

  dfs(0, 0);

  for (int i = 0; i < n; ++i) {
    int sz = n - colors[0][i];
    ans[i] -= 1LL * sz * (sz + 1) / 2;
  }

  for (int i = 0; i < n; ++i) {
    printf("%lld\n", ans[i]);
  }
  return 0;
}

### Prompt

Please provide a Cpp coded solution 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 = 200005;

int n;
int c[N];
vector<int> edges[N];

long long ans[N];

int sz[N];
map<int, int> colors[N];
void dfs(int u, int p) {
  sz[u] = 1;
  int heavy = u;
  for (int v: edges[u]) {
    if (v == p) continue;
    dfs(v, u);

    sz[u] += sz[v];
    if (heavy == u || colors[heavy].size() < colors[v].size()) {
      heavy = v;
    }

    int comp = sz[v] - colors[v][c[u]];
    ans[c[u]] -= 1LL * comp * (comp + 1) / 2;
    colors[v].erase(c[u]);
  }

  swap(colors[u], colors[heavy]);

  for (int v: edges[u]) {
    if (v == p) continue;

    for (pair<int, int> p: colors[v]) {
      colors[u][p.first] += p.second;
    }
  }

  colors[u][c[u]] += sz[u];
}

int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; ++i) {
    scanf("%d", &c[i]);
    --c[i];
    ans[i] = 1LL * n * (n+1) / 2;
  }
  for (int i = 1; i < n; ++i) {
    int u, v;
    scanf("%d %d", &u, &v);
    --u, --v;

    edges[u].push_back(v);
    edges[v].push_back(u);
  }

  dfs(0, 0);

  for (int i = 0; i < n; ++i) {
    int sz = n - colors[0][i];
    ans[i] -= 1LL * sz * (sz + 1) / 2;
  }

  for (int i = 0; i < n; ++i) {
    printf("%lld\n", ans[i]);
  }
  return 0;
}

```