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

const ll MAX = 2e5;
ll siz[MAX], cnt[MAX], ans[MAX], color[MAX];
vector<ll> g[MAX];
void dfs1(ll from, ll at){
    siz[at]++;
    for(ll i : g[at]) if(i != from){
        dfs1(at, i);
        siz[at] += siz[i];
    }
}
void dfs2(ll from, ll at){
    const ll c = color[at];
    ll tmp = cnt[c];
    for(ll i : g[at]) if(i != from){
        cnt[c] = siz[i];
        dfs2(at, i);
        ans[c] -= cnt[c] * (cnt[c] + 1) / 2;
    }
    cnt[c] = tmp - siz[at];
}
int main(){
    ll n;
    cin >> n;
    for(ll i = 0; i < n; i++){
        ll c;
        cin >> c;
        c--;
        color[i] = c;
        cnt[c] = n;
        ans[c] = n * (n + 1) / 2;
    }
    for(ll i = 1; i < n; i++){
        ll a, b;
        cin >> a >> b;
        a--; b--;
        g[a].push_back(b);
        g[b].push_back(a);
    }
    dfs1(-1, 0);
    dfs2(-1, 0);
    for(ll i = 0; i < n; i++) ans[i] -= cnt[i] * (cnt[i] + 1) / 2;
    for(ll i = 0; i < n; i++) cout << ans[i] << ' ';
}
### Prompt

In Cpp, your task is to solve 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;

const ll MAX = 2e5;
ll siz[MAX], cnt[MAX], ans[MAX], color[MAX];
vector<ll> g[MAX];
void dfs1(ll from, ll at){
    siz[at]++;
    for(ll i : g[at]) if(i != from){
        dfs1(at, i);
        siz[at] += siz[i];
    }
}
void dfs2(ll from, ll at){
    const ll c = color[at];
    ll tmp = cnt[c];
    for(ll i : g[at]) if(i != from){
        cnt[c] = siz[i];
        dfs2(at, i);
        ans[c] -= cnt[c] * (cnt[c] + 1) / 2;
    }
    cnt[c] = tmp - siz[at];
}
int main(){
    ll n;
    cin >> n;
    for(ll i = 0; i < n; i++){
        ll c;
        cin >> c;
        c--;
        color[i] = c;
        cnt[c] = n;
        ans[c] = n * (n + 1) / 2;
    }
    for(ll i = 1; i < n; i++){
        ll a, b;
        cin >> a >> b;
        a--; b--;
        g[a].push_back(b);
        g[b].push_back(a);
    }
    dfs1(-1, 0);
    dfs2(-1, 0);
    for(ll i = 0; i < n; i++) ans[i] -= cnt[i] * (cnt[i] + 1) / 2;
    for(ll i = 0; i < n; i++) cout << ans[i] << ' ';
}
```