#include<bits/stdc++.h>
using namespace std;
const int MX = 2e5+5;

typedef long long ll;

int color[MX];
vector<int> adj[MX];
ll dp[MX], depth[MX], overc[MX];

ll sum(ll n) { return n*(n+1)/2; }

void dfs(int u, int par) {
    depth[u] = 1; ll cur = dp[color[u]];
    for(auto& v: adj[u]) if (v != par) {
        dp[color[u]] = 0; dfs(v,u);
        depth[u] += depth[v];
        overc[color[u]] += sum(depth[v]-dp[color[u]]);
    }
    dp[color[u]] = cur+depth[u];
}

int32_t main() {

    ios_base::sync_with_stdio(0); cin.tie(0);

    int N; cin >> N;

    for (int i = 0; i < N; ++i) {
        cin >> color[i];
    }

    for (int i = 0; i < N-1; ++i) {
        int u,v; cin >> u >> v; --u,--v;
        adj[u].push_back(v);
        adj[v].push_back(u);
    }

    dfs(0,-1);

    for (int i = 1; i <= N; ++i) {
        cout << sum(N)-sum(N-dp[i])-overc[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;
const int MX = 2e5+5;

typedef long long ll;

int color[MX];
vector<int> adj[MX];
ll dp[MX], depth[MX], overc[MX];

ll sum(ll n) { return n*(n+1)/2; }

void dfs(int u, int par) {
    depth[u] = 1; ll cur = dp[color[u]];
    for(auto& v: adj[u]) if (v != par) {
        dp[color[u]] = 0; dfs(v,u);
        depth[u] += depth[v];
        overc[color[u]] += sum(depth[v]-dp[color[u]]);
    }
    dp[color[u]] = cur+depth[u];
}

int32_t main() {

    ios_base::sync_with_stdio(0); cin.tie(0);

    int N; cin >> N;

    for (int i = 0; i < N; ++i) {
        cin >> color[i];
    }

    for (int i = 0; i < N-1; ++i) {
        int u,v; cin >> u >> v; --u,--v;
        adj[u].push_back(v);
        adj[v].push_back(u);
    }

    dfs(0,-1);

    for (int i = 1; i <= N; ++i) {
        cout << sum(N)-sum(N-dp[i])-overc[i] << "\n";
    }

}

```