#include <iostream>
#include <algorithm>
#include <vector>
#define rip(i, n, s) for (int i = (s);i < (int)( n ); i++)
typedef long long ll;
using namespace std;
const int MAX_N = 200200;
int c[MAX_N];
ll ans[MAX_N], lam[MAX_N], co[MAX_N];
ll cal(ll x){
    return x*(x+1)/2;
}
int dfs2(vector<vector<int>> &da, vector<bool> &used, int now){
    ll tmp = lam[c[now]];
    lam[c[now]] = 0;
    rip(i, da[now].size(),0){
        if(used[da[now][i]]) continue;
        used[da[now][i]] = true;
        co[now] += dfs2(da, used, da[now][i]);
        ans[c[now]] += cal(co[da[now][i]]-lam[c[now]]);
        lam[c[now]] = 0;
    }
    ++co[now];
    lam[c[now]] = tmp + co[now];
    return co[now];
}
int main(){
    ll n; cin >> n;
    rip(i,n,0) cin >> c[i], --c[i];
    vector<vector<int>> da(n, vector<int>(0));
    rip(i,n-1,0){
        int a,b; cin >> a >> b;
        --a;--b;
        da[a].push_back(b);
        da[b].push_back(a);
    }
    vector<bool> used(n, false);
    used[0] = true;
    dfs2(da, used,0);
    rip(i,n,0){
        ans[i] += cal(n - lam[i]);
    }
    rip(i,n,0){
        printf("%lld\n", n*(n+1)/2 - ans[i]);
    }
}

### 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 <iostream>
#include <algorithm>
#include <vector>
#define rip(i, n, s) for (int i = (s);i < (int)( n ); i++)
typedef long long ll;
using namespace std;
const int MAX_N = 200200;
int c[MAX_N];
ll ans[MAX_N], lam[MAX_N], co[MAX_N];
ll cal(ll x){
    return x*(x+1)/2;
}
int dfs2(vector<vector<int>> &da, vector<bool> &used, int now){
    ll tmp = lam[c[now]];
    lam[c[now]] = 0;
    rip(i, da[now].size(),0){
        if(used[da[now][i]]) continue;
        used[da[now][i]] = true;
        co[now] += dfs2(da, used, da[now][i]);
        ans[c[now]] += cal(co[da[now][i]]-lam[c[now]]);
        lam[c[now]] = 0;
    }
    ++co[now];
    lam[c[now]] = tmp + co[now];
    return co[now];
}
int main(){
    ll n; cin >> n;
    rip(i,n,0) cin >> c[i], --c[i];
    vector<vector<int>> da(n, vector<int>(0));
    rip(i,n-1,0){
        int a,b; cin >> a >> b;
        --a;--b;
        da[a].push_back(b);
        da[b].push_back(a);
    }
    vector<bool> used(n, false);
    used[0] = true;
    dfs2(da, used,0);
    rip(i,n,0){
        ans[i] += cal(n - lam[i]);
    }
    rip(i,n,0){
        printf("%lld\n", n*(n+1)/2 - ans[i]);
    }
}

```