#include <bits/stdc++.h>
using namespace std;
using ll  = long long;
using vl  = vector<ll>;
using vvl = vector<vl>;
#define rep(i,n) for(int i = 0; i < (n); ++i)
const int MAX_N = 2e5+7;

ll N;
vvl G(MAX_N);
vl cs(MAX_N), tree_size(MAX_N), color_cnts(MAX_N,0);
vl ans(MAX_N);

ll nc2(ll x) { return x*(x-1)/2; }

void calc_size(int v, int par = -1) {
    tree_size[v] = 1;
    for (auto to : G[v]) {
        if (to == par) continue;
        calc_size(to, v);
        tree_size[v] += tree_size[to];
    }
}
void dfs(int v, int par = -1) {
    ll prev_cnt = color_cnts[cs[v]];
    for (auto to : G[v]) {
        if (to == par) continue;
        ll prev = color_cnts[cs[v]];
        dfs(to, v);
        ll diff = color_cnts[cs[v]] - prev;
        ll cnt = tree_size[to] - diff;
        // cerr << v << "->" << to << ": " << cnt << endl;
        ans[cs[v]] -= nc2(cnt);
    }
    color_cnts[cs[v]] = prev_cnt + tree_size[v];
}

int main() {
    cin >> N;
    ans.assign(MAX_N, nc2(N));
    rep(i,N) cin >> cs[i+1], ans[cs[i+1]]++;
    rep(i,N-1) {
        int u, v; cin >> u >> v;
        G[u].push_back(v);
        G[v].push_back(u);
    }
    calc_size(1);
    dfs(1);
    rep(c,N) ans[c+1] -= nc2(N-color_cnts[c+1]);
    rep(i,N) cout << ans[i+1] << "\n";
    return 0;
}

### Prompt

Construct a Cpp code solution to the problem outlined:

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;
using vl  = vector<ll>;
using vvl = vector<vl>;
#define rep(i,n) for(int i = 0; i < (n); ++i)
const int MAX_N = 2e5+7;

ll N;
vvl G(MAX_N);
vl cs(MAX_N), tree_size(MAX_N), color_cnts(MAX_N,0);
vl ans(MAX_N);

ll nc2(ll x) { return x*(x-1)/2; }

void calc_size(int v, int par = -1) {
    tree_size[v] = 1;
    for (auto to : G[v]) {
        if (to == par) continue;
        calc_size(to, v);
        tree_size[v] += tree_size[to];
    }
}
void dfs(int v, int par = -1) {
    ll prev_cnt = color_cnts[cs[v]];
    for (auto to : G[v]) {
        if (to == par) continue;
        ll prev = color_cnts[cs[v]];
        dfs(to, v);
        ll diff = color_cnts[cs[v]] - prev;
        ll cnt = tree_size[to] - diff;
        // cerr << v << "->" << to << ": " << cnt << endl;
        ans[cs[v]] -= nc2(cnt);
    }
    color_cnts[cs[v]] = prev_cnt + tree_size[v];
}

int main() {
    cin >> N;
    ans.assign(MAX_N, nc2(N));
    rep(i,N) cin >> cs[i+1], ans[cs[i+1]]++;
    rep(i,N-1) {
        int u, v; cin >> u >> v;
        G[u].push_back(v);
        G[v].push_back(u);
    }
    calc_size(1);
    dfs(1);
    rep(c,N) ans[c+1] -= nc2(N-color_cnts[c+1]);
    rep(i,N) cout << ans[i+1] << "\n";
    return 0;
}

```