#include"bits/stdc++.h"
#include<unordered_set>
#include<unordered_map>
#include<random>
using namespace std;
typedef long long ll;
const ll MOD = (ll)(1e9+7);
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(), (x).end()
#define rep(i, n) for(int (i)=0; (i)<(int)(n); (i)++)
int dx[4]={ 1,0,-1,0 };
int dy[4]={ 0,1,0,-1 };

int N;
int c[200000];
vector<int> e[200000];

ll ans[200000] ={};
int used[200000] ={};

ll calc(ll n){
    return n*(n-1)/2 + n;
}

int dfs(int n, int p){
    int usedMemo = used[c[n]];
    int siz = 1;
    rep(i, e[n].size()){
        if(e[n][i] == p) continue;
        used[c[n]] = 0;
        int tmp = dfs(e[n][i], n);
        siz += tmp;
        ans[c[n]] += calc(tmp-used[c[n]]);
    }
    usedMemo += siz;
    used[c[n]] = usedMemo;
    return siz;
}

signed main(){
    cin >> N;
    rep(i, N){
        cin >> c[i];
        c[i]--;
    }
    rep(i, N-1){
        int a, b;
        cin >> a >> b;
        a--; b--;
        e[a].pb(b);
        e[b].pb(a);
    }
    dfs(0, -1);

    rep(i, N){
        cout << calc(N) - (ans[i] + calc(N-used[i])) << endl;
    }
}
### Prompt

Develop a solution in Cpp to the problem described below:

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"
#include<unordered_set>
#include<unordered_map>
#include<random>
using namespace std;
typedef long long ll;
const ll MOD = (ll)(1e9+7);
#define pb push_back
#define mp make_pair
#define all(x) (x).begin(), (x).end()
#define rep(i, n) for(int (i)=0; (i)<(int)(n); (i)++)
int dx[4]={ 1,0,-1,0 };
int dy[4]={ 0,1,0,-1 };

int N;
int c[200000];
vector<int> e[200000];

ll ans[200000] ={};
int used[200000] ={};

ll calc(ll n){
    return n*(n-1)/2 + n;
}

int dfs(int n, int p){
    int usedMemo = used[c[n]];
    int siz = 1;
    rep(i, e[n].size()){
        if(e[n][i] == p) continue;
        used[c[n]] = 0;
        int tmp = dfs(e[n][i], n);
        siz += tmp;
        ans[c[n]] += calc(tmp-used[c[n]]);
    }
    usedMemo += siz;
    used[c[n]] = usedMemo;
    return siz;
}

signed main(){
    cin >> N;
    rep(i, N){
        cin >> c[i];
        c[i]--;
    }
    rep(i, N-1){
        int a, b;
        cin >> a >> b;
        a--; b--;
        e[a].pb(b);
        e[b].pb(a);
    }
    dfs(0, -1);

    rep(i, N){
        cout << calc(N) - (ans[i] + calc(N-used[i])) << endl;
    }
}
```