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

constexpr int MAX = 200010;

// 関数にしておくと便利
ll subcalc(ll x) { return x * (x + 1) / 2; }

int n;
int c[MAX];
vector<int> G[MAX];

int sz[MAX];
stack<int> last[MAX];
ll ans[MAX];

// 部分木のサイズを求める
void dfs1(int v, int p) {
    sz[v] = 1;
    for(int u : G[v]) {
        if(u != p) {
            dfs1(u, v);
            sz[v] += sz[u];
        }
    }
}

// 答えの計算
void dfs2(int v, int p) {
    if(v == 0) { // 根の上の頂点 (?) で切断しておく
        for(int i = 1; i <= n; i++)
            last[i].push(n);
        dfs2(1, 0);
        for(int i = 1; i <= n; i++)
            ans[i] += subcalc(last[i].top());
    } else {
        last[c[v]].top() -= sz[v];

        for(int u : G[v]) {
            if(u != p) {
                last[c[v]].push(sz[u]);
                dfs2(u, v);
                ans[c[v]] += subcalc(last[c[v]].top());
                last[c[v]].pop();
            }
        }
    }
}

int main() {
    cin >> n;
    for(int i = 1; i <= n; i++) {
        cin >> c[i];
        last[i].push(n);
    }
    for(int i = 0; i < n - 1; i++) {
        int a, b;
        cin >> a >> b;
        G[a].push_back(b);
        G[b].push_back(a);
    }

    dfs1(1, 0);
    dfs2(0, 0);

    for(int i = 1; i <= n; i++)
        cout << subcalc(n) - ans[i] << endl;

    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>
using namespace std;
using ll = long long;

constexpr int MAX = 200010;

// 関数にしておくと便利
ll subcalc(ll x) { return x * (x + 1) / 2; }

int n;
int c[MAX];
vector<int> G[MAX];

int sz[MAX];
stack<int> last[MAX];
ll ans[MAX];

// 部分木のサイズを求める
void dfs1(int v, int p) {
    sz[v] = 1;
    for(int u : G[v]) {
        if(u != p) {
            dfs1(u, v);
            sz[v] += sz[u];
        }
    }
}

// 答えの計算
void dfs2(int v, int p) {
    if(v == 0) { // 根の上の頂点 (?) で切断しておく
        for(int i = 1; i <= n; i++)
            last[i].push(n);
        dfs2(1, 0);
        for(int i = 1; i <= n; i++)
            ans[i] += subcalc(last[i].top());
    } else {
        last[c[v]].top() -= sz[v];

        for(int u : G[v]) {
            if(u != p) {
                last[c[v]].push(sz[u]);
                dfs2(u, v);
                ans[c[v]] += subcalc(last[c[v]].top());
                last[c[v]].pop();
            }
        }
    }
}

int main() {
    cin >> n;
    for(int i = 1; i <= n; i++) {
        cin >> c[i];
        last[i].push(n);
    }
    for(int i = 0; i < n - 1; i++) {
        int a, b;
        cin >> a >> b;
        G[a].push_back(b);
        G[b].push_back(a);
    }

    dfs1(1, 0);
    dfs2(0, 0);

    for(int i = 1; i <= n; i++)
        cout << subcalc(n) - ans[i] << endl;

    return 0;
}
```