#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define N 200005
#define int long long
#define il inline
#define rep(i,a,n) for (int i=a;i<=n;i++)
#define per(i,n,a) for(int i=n;i>=a;i--)
#define  pi  acos(-1)
#define ld  double
#define mod 1000000007
#pragma GCC optimize(2)
#define IOS ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
int c[N];
vector<ll> g[N],ans;
int sz[N],cnt[N];
void dfs1(int x,int fa){
    sz[x]=1;
    for(auto i:g[x]){
        if(i!=fa){
            dfs1(i,x);
            sz[x]+=sz[i];
        }
    }
}
void dfs2(int x,int fa){
    int tmp=cnt[c[x]];
    for(auto i:g[x]){
        if(i==fa)continue;
        int now=cnt[c[x]];
        dfs2(i,x);
        int dif=cnt[c[x]]-now;
        int d=sz[i]-dif;
        ans[c[x]]-=(d-1)*d/2;
    }
    cnt[c[x]]=tmp+sz[x];
}
signed main() {
   // IOS
    int n;
    cin>>n;
    memset(cnt,0, sizeof(cnt));
    ans.assign(n+10,n*(n-1)/2);
    for(int i=1;i<=n;i++){cin>>c[i];ans[c[i]]++;}
    for (int j = 0; j <n-1 ; ++j) {
        int u,v;
        cin>>u>>v;
        g[v].push_back(u);
        g[u].push_back(v);
    }
    dfs1(1,0);
    dfs2(1,0);
    for (int i = 1; i <=n ; ++i) {
        ans[i]-=(n-cnt[i])*(n-cnt[i]-1)/2;
    }
    for (int i = 1; i <=n ; ++i) {
        cout<<ans[i]<<endl;
    }
    return 0;
}
### Prompt

Please provide a cpp coded solution 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>
using namespace std;
#define ll long long
#define N 200005
#define int long long
#define il inline
#define rep(i,a,n) for (int i=a;i<=n;i++)
#define per(i,n,a) for(int i=n;i>=a;i--)
#define  pi  acos(-1)
#define ld  double
#define mod 1000000007
#pragma GCC optimize(2)
#define IOS ios_base::sync_with_stdio(false);cin.tie(0);cout.tie(0);
int c[N];
vector<ll> g[N],ans;
int sz[N],cnt[N];
void dfs1(int x,int fa){
    sz[x]=1;
    for(auto i:g[x]){
        if(i!=fa){
            dfs1(i,x);
            sz[x]+=sz[i];
        }
    }
}
void dfs2(int x,int fa){
    int tmp=cnt[c[x]];
    for(auto i:g[x]){
        if(i==fa)continue;
        int now=cnt[c[x]];
        dfs2(i,x);
        int dif=cnt[c[x]]-now;
        int d=sz[i]-dif;
        ans[c[x]]-=(d-1)*d/2;
    }
    cnt[c[x]]=tmp+sz[x];
}
signed main() {
   // IOS
    int n;
    cin>>n;
    memset(cnt,0, sizeof(cnt));
    ans.assign(n+10,n*(n-1)/2);
    for(int i=1;i<=n;i++){cin>>c[i];ans[c[i]]++;}
    for (int j = 0; j <n-1 ; ++j) {
        int u,v;
        cin>>u>>v;
        g[v].push_back(u);
        g[u].push_back(v);
    }
    dfs1(1,0);
    dfs2(1,0);
    for (int i = 1; i <=n ; ++i) {
        ans[i]-=(n-cnt[i])*(n-cnt[i]-1)/2;
    }
    for (int i = 1; i <=n ; ++i) {
        cout<<ans[i]<<endl;
    }
    return 0;
}
```