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

using ll  = long long;
using vl  = vector<ll>;
using vvl = vector<vl>;
#define rng(i,a,b) for(int i = (a); i < (b); ++i)
#define rep(i,n) for(int i = 0; i < (n); ++i)

ll N, sum = 0;
vvl edges;
vl color, ans, cnts, sub_size;

void dfs(int v, int parent) {
    sub_size[v] = 1;
    ll prev_dfs = cnts[color[v]];
    for (auto to : edges[v]) {
        if (to == parent) continue;
        ll prev = sum - cnts[color[v]];
        dfs(to, v);
        sub_size[v] += sub_size[to];
        ll next = sum - cnts[color[v]];
        ll diff = next - prev;
        ans[color[v]] -= diff * (diff-1) / 2;
    }
    cnts[color[v]] = prev_dfs + sub_size[v];
    sum++;
}

int main() {
    cin >> N;
    edges.resize(N);
    color.resize(N);
    sub_size.resize(N);
    ans.assign(N,N*(N-1)/2);
    cnts.assign(N,0);
    rep(i,N) { cin >> color[i]; color[i]--; }
    rep(i,N-1) {
        int a, b; cin >> a >> b;
        a--; b--;
        edges[a].push_back(b);
        edges[b].push_back(a);
    }
    dfs(0,-1);
    rep(c,N) {
        ans[c] -= (N-cnts[c]) * (N-cnts[c]-1) / 2;
    }
    rep(i,N) ans[color[i]]++;
    rep(i,N) cout << ans[i] << endl;
    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;

using ll  = long long;
using vl  = vector<ll>;
using vvl = vector<vl>;
#define rng(i,a,b) for(int i = (a); i < (b); ++i)
#define rep(i,n) for(int i = 0; i < (n); ++i)

ll N, sum = 0;
vvl edges;
vl color, ans, cnts, sub_size;

void dfs(int v, int parent) {
    sub_size[v] = 1;
    ll prev_dfs = cnts[color[v]];
    for (auto to : edges[v]) {
        if (to == parent) continue;
        ll prev = sum - cnts[color[v]];
        dfs(to, v);
        sub_size[v] += sub_size[to];
        ll next = sum - cnts[color[v]];
        ll diff = next - prev;
        ans[color[v]] -= diff * (diff-1) / 2;
    }
    cnts[color[v]] = prev_dfs + sub_size[v];
    sum++;
}

int main() {
    cin >> N;
    edges.resize(N);
    color.resize(N);
    sub_size.resize(N);
    ans.assign(N,N*(N-1)/2);
    cnts.assign(N,0);
    rep(i,N) { cin >> color[i]; color[i]--; }
    rep(i,N-1) {
        int a, b; cin >> a >> b;
        a--; b--;
        edges[a].push_back(b);
        edges[b].push_back(a);
    }
    dfs(0,-1);
    rep(c,N) {
        ans[c] -= (N-cnts[c]) * (N-cnts[c]-1) / 2;
    }
    rep(i,N) ans[color[i]]++;
    rep(i,N) cout << ans[i] << endl;
    return 0;
}

```