#include <bits/stdc++.h>

using namespace std;
const int N = 2e5 + 5;
int sz[N], c[N], lst[N], h[N];
long long ans[N], f[N], val[N], s[N];
vector <int> g[N];

void dfs(int v, int p = 0) {
    int prv = lst[c[v]];
    for (int u : g[v]) if (u != p) lst[c[v]] = u, dfs(u, v), ans[c[v]] += 1ll * sz[v] * sz[u], sz[v] += sz[u];
    lst[c[v]] = prv, sz[v]++;
    if (prv > 1) val[prv] += sz[v];
    else s[c[v]] += sz[v];
    return ;
}
void Solve(int v, int p = 0) {
    int prv = lst[c[v]];
    ans[c[v]] += 1ll * sz[v] * (sz[prv] - sz[v] + 1);
    if (prv > 1) f[c[v]] += 1ll * sz[v] * (val[prv] - sz[v]);
    else f[c[v]] += 1ll * sz[v] * (s[c[v]] - sz[v]);
    for (int u : g[v]) if (u != p) lst[c[v]] = u, Solve(u, v);
    lst[c[v]] = prv;
}

int main() {
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int n; cin >> n;
    for (int i = 1; i <= n; i++) cin >> c[i];
    for (int i = 0; i < n - 1; i++) {
        int u, v; cin >> u >> v;
        g[u].push_back(v), g[v].push_back(u);
    }
    fill(lst, lst + N, 1), dfs(1), fill(lst, lst + N, 1), Solve(1);
    for (int i = 1; i <= n; i++) cout << ans[i] - (f[i]/2) << "\n";


    return 0;
}

### 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;
const int N = 2e5 + 5;
int sz[N], c[N], lst[N], h[N];
long long ans[N], f[N], val[N], s[N];
vector <int> g[N];

void dfs(int v, int p = 0) {
    int prv = lst[c[v]];
    for (int u : g[v]) if (u != p) lst[c[v]] = u, dfs(u, v), ans[c[v]] += 1ll * sz[v] * sz[u], sz[v] += sz[u];
    lst[c[v]] = prv, sz[v]++;
    if (prv > 1) val[prv] += sz[v];
    else s[c[v]] += sz[v];
    return ;
}
void Solve(int v, int p = 0) {
    int prv = lst[c[v]];
    ans[c[v]] += 1ll * sz[v] * (sz[prv] - sz[v] + 1);
    if (prv > 1) f[c[v]] += 1ll * sz[v] * (val[prv] - sz[v]);
    else f[c[v]] += 1ll * sz[v] * (s[c[v]] - sz[v]);
    for (int u : g[v]) if (u != p) lst[c[v]] = u, Solve(u, v);
    lst[c[v]] = prv;
}

int main() {
    ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
    int n; cin >> n;
    for (int i = 1; i <= n; i++) cin >> c[i];
    for (int i = 0; i < n - 1; i++) {
        int u, v; cin >> u >> v;
        g[u].push_back(v), g[v].push_back(u);
    }
    fill(lst, lst + N, 1), dfs(1), fill(lst, lst + N, 1), Solve(1);
    for (int i = 1; i <= n; i++) cout << ans[i] - (f[i]/2) << "\n";


    return 0;
}

```