#include <iostream>
#include <vector>
using i64 = long long;

std::vector<std::vector<int>> g, q;
std::vector<std::vector<i64>> comp;
std::vector<int> c, sz;

int szdfs(int v, int p) {
    for (const int to : g[v]) {
        if (to == p) continue;
        sz[v] += szdfs(to, v);
    }
    return sz[v];
}

void dfs(int v, int p) {
    const int j = c[v];
    comp[j][q[j].back()] -= sz[v];
    for (const int to : g[v]) {
        if (to == p) continue;
        q[j].push_back(comp[j].size());
        comp[j].push_back(sz[to]);
        dfs(to, v);
        q[j].pop_back();
    }
}

int main() {
    i64 n;
    std::cin >> n;
    const i64 all = n * (n + 1) / 2;
    g.resize(n); c.resize(n); q.resize(n); comp.resize(n); sz.resize(n, 1);
    for (auto &e : comp) e.push_back(n);
    for (auto &e : q) e.push_back(0);
    for (auto &e : c) {
        std::cin >> e;
        --e;
    }
    for (int i = 1; i < n; i++) {
        int u, v;
        std::cin >> u >> v;
        --u; --v;
        g[u].push_back(v);
        g[v].push_back(u);
    }

    szdfs(0, -1);
    dfs(0, -1);

    for (int i = 0; i < n; i++) {
        i64 s = 0;
        for (const i64 r : comp[i]) s += r * (r + 1) / 2;
        std::cout << all - s << '\n';
    }

    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 <iostream>
#include <vector>
using i64 = long long;

std::vector<std::vector<int>> g, q;
std::vector<std::vector<i64>> comp;
std::vector<int> c, sz;

int szdfs(int v, int p) {
    for (const int to : g[v]) {
        if (to == p) continue;
        sz[v] += szdfs(to, v);
    }
    return sz[v];
}

void dfs(int v, int p) {
    const int j = c[v];
    comp[j][q[j].back()] -= sz[v];
    for (const int to : g[v]) {
        if (to == p) continue;
        q[j].push_back(comp[j].size());
        comp[j].push_back(sz[to]);
        dfs(to, v);
        q[j].pop_back();
    }
}

int main() {
    i64 n;
    std::cin >> n;
    const i64 all = n * (n + 1) / 2;
    g.resize(n); c.resize(n); q.resize(n); comp.resize(n); sz.resize(n, 1);
    for (auto &e : comp) e.push_back(n);
    for (auto &e : q) e.push_back(0);
    for (auto &e : c) {
        std::cin >> e;
        --e;
    }
    for (int i = 1; i < n; i++) {
        int u, v;
        std::cin >> u >> v;
        --u; --v;
        g[u].push_back(v);
        g[v].push_back(u);
    }

    szdfs(0, -1);
    dfs(0, -1);

    for (int i = 0; i < n; i++) {
        i64 s = 0;
        for (const i64 r : comp[i]) s += r * (r + 1) / 2;
        std::cout << all - s << '\n';
    }

    return 0;
}

```