#include <cstdio>
#include <vector>

using namespace std;

const int MAXN=2e5+10;

typedef long long int ll;

int color[MAXN];
ll N;
vector<int> graph[MAXN];

ll sizes[MAXN];
ll cnt[MAXN];
ll ans[MAXN];

void DFS(int cur,int parent){
    sizes[cur]=1;
    ll temp=cnt[color[cur]];

    for(auto adj:graph[cur]){
        if (adj==parent) continue;
        ll temp=cnt[color[cur]];
        DFS(adj,cur);
        sizes[cur]+=sizes[adj];
        ll diff=cnt[color[cur]]-temp;
        ll sz=sizes[adj]-diff;
        ans[color[cur]]-=sz*(sz-1)/2;
    }

    cnt[color[cur]]=temp+sizes[cur];

}

int main(){
    scanf("%lld",&N);
    for(int i=1;i<=N;i++) scanf("%d",color+i),ans[i]=N*(N-1)/2;

    for(int i=1;i<=N-1;i++){
        int s,e;
        scanf("%d%d",&s,&e);
        graph[s].push_back(e);
        graph[e].push_back(s);
    }

    DFS(1,1);

    for(int c=1;c<=N;c++){
        ll nxt=N-cnt[c];
        ans[c]-=nxt*(nxt-1)/2;
    }

    for(int i=1;i<=N;i++) ans[color[i]]++;

    for(int c=1;c<=N;c++) printf("%lld\n",ans[c]);

    return 0;
}
### Prompt

Please formulate a CPP solution to 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 <cstdio>
#include <vector>

using namespace std;

const int MAXN=2e5+10;

typedef long long int ll;

int color[MAXN];
ll N;
vector<int> graph[MAXN];

ll sizes[MAXN];
ll cnt[MAXN];
ll ans[MAXN];

void DFS(int cur,int parent){
    sizes[cur]=1;
    ll temp=cnt[color[cur]];

    for(auto adj:graph[cur]){
        if (adj==parent) continue;
        ll temp=cnt[color[cur]];
        DFS(adj,cur);
        sizes[cur]+=sizes[adj];
        ll diff=cnt[color[cur]]-temp;
        ll sz=sizes[adj]-diff;
        ans[color[cur]]-=sz*(sz-1)/2;
    }

    cnt[color[cur]]=temp+sizes[cur];

}

int main(){
    scanf("%lld",&N);
    for(int i=1;i<=N;i++) scanf("%d",color+i),ans[i]=N*(N-1)/2;

    for(int i=1;i<=N-1;i++){
        int s,e;
        scanf("%d%d",&s,&e);
        graph[s].push_back(e);
        graph[e].push_back(s);
    }

    DFS(1,1);

    for(int c=1;c<=N;c++){
        ll nxt=N-cnt[c];
        ans[c]-=nxt*(nxt-1)/2;
    }

    for(int i=1;i<=N;i++) ans[color[i]]++;

    for(int c=1;c<=N;c++) printf("%lld\n",ans[c]);

    return 0;
}
```