#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MAX_N=2e5+10;

ll N;
int c[MAX_N];
ll sub[MAX_N],ans[MAX_N],cnt[MAX_N];
vector<int> G[MAX_N];

void dfs(int v,int p){
    sub[v]=1;
    ll pre=cnt[c[v]];
    for (int u:G[v]){
        if (u==p) continue;
        cnt[c[v]]=0;
        dfs(u,v);
        sub[v]+=sub[u];
        ans[c[v]]+=(sub[u]-cnt[c[v]])*(sub[u]-cnt[c[v]]+1)/2;
    }
    cnt[c[v]]=pre+sub[v];
}

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    cin >> N;
    for (int i=0;i<N;++i) cin >> c[i],--c[i];
    for (int i=0;i<N-1;++i){
        int a,b; cin >> a >> b; --a,--b;
        G[a].emplace_back(b);
        G[b].emplace_back(a);
    }
    dfs(0,-1);
    for (int i=0;i<N;++i) cout << N*(N+1)/2-ans[i]-(N-cnt[i])*(N-cnt[i]+1)/2 << '\n';
}
### 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;
typedef long long ll;
const int MAX_N=2e5+10;

ll N;
int c[MAX_N];
ll sub[MAX_N],ans[MAX_N],cnt[MAX_N];
vector<int> G[MAX_N];

void dfs(int v,int p){
    sub[v]=1;
    ll pre=cnt[c[v]];
    for (int u:G[v]){
        if (u==p) continue;
        cnt[c[v]]=0;
        dfs(u,v);
        sub[v]+=sub[u];
        ans[c[v]]+=(sub[u]-cnt[c[v]])*(sub[u]-cnt[c[v]]+1)/2;
    }
    cnt[c[v]]=pre+sub[v];
}

int main(){
    cin.tie(0);
    ios::sync_with_stdio(false);
    cin >> N;
    for (int i=0;i<N;++i) cin >> c[i],--c[i];
    for (int i=0;i<N-1;++i){
        int a,b; cin >> a >> b; --a,--b;
        G[a].emplace_back(b);
        G[b].emplace_back(a);
    }
    dfs(0,-1);
    for (int i=0;i<N;++i) cout << N*(N+1)/2-ans[i]-(N-cnt[i])*(N-cnt[i]+1)/2 << '\n';
}
```