#include <iostream>
#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9+7;
const long long MAXN = 2e5+5;
long long n;
vector<long long> v1[MAXN];
long long ans[MAXN];
long long res[MAXN];
long long col[MAXN];

long long sz[MAXN];
long long f(long long x){
    return 1LL*x*(x+1)/2;
}
void dfs(long long curr, long long par){
    long long tmp = res[col[curr]];
    for(long long x:v1[curr]){
        if(x!=par){
            res[col[curr]] = 0;
            dfs(x,curr);
            sz[curr] += sz[x];
            ans[col[curr]]-=f(sz[x]-res[col[curr]]);
        }
    }
    res[col[curr]] = tmp+sz[curr];
}
int main() {
    cin>>n;
    for(long long i=1;i<=n;i++){
        cin>>col[i];
        sz[i] = 1;
    }
    for(long long i=1;i<n;i++){
        long long x,y;
        cin>>x>>y;
        v1[x].push_back(y);
        v1[y].push_back(x);
    }
    dfs(1,1);
    for(long long i=1;i<=n;i++){
        //cout<<col[i]<<endl;
        cout<<(ans[i]+f(n)-f(n-res[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 <iostream>
#include <bits/stdc++.h>
using namespace std;
const long long MOD = 1e9+7;
const long long MAXN = 2e5+5;
long long n;
vector<long long> v1[MAXN];
long long ans[MAXN];
long long res[MAXN];
long long col[MAXN];

long long sz[MAXN];
long long f(long long x){
    return 1LL*x*(x+1)/2;
}
void dfs(long long curr, long long par){
    long long tmp = res[col[curr]];
    for(long long x:v1[curr]){
        if(x!=par){
            res[col[curr]] = 0;
            dfs(x,curr);
            sz[curr] += sz[x];
            ans[col[curr]]-=f(sz[x]-res[col[curr]]);
        }
    }
    res[col[curr]] = tmp+sz[curr];
}
int main() {
    cin>>n;
    for(long long i=1;i<=n;i++){
        cin>>col[i];
        sz[i] = 1;
    }
    for(long long i=1;i<n;i++){
        long long x,y;
        cin>>x>>y;
        v1[x].push_back(y);
        v1[y].push_back(x);
    }
    dfs(1,1);
    for(long long i=1;i<=n;i++){
        //cout<<col[i]<<endl;
        cout<<(ans[i]+f(n)-f(n-res[i]))<<endl;
    }
}
```