#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;

const int N=2e5+5;
int he[N],ver[N*2],nex[N*2],tot;
void addedge(int,int);
int c[N];
int sum[N],siz[N];
ll ans[N];
void dfs(int,int);

void init();
void work();
int ca=1;
int main(){
//    freopen("in.txt","r",stdin);
    ios::sync_with_stdio(0);
    cin.tie(0),cout.tie(0);
//    init();
//    int T;cin>>T;
//    while(T--)
    work();
    return 0;
}
void work(){
    int n;cin>>n;
    for(int i=1;i<=n;i++) cin>>c[i];
    int u,v;tot=1;
    for(int i=1;i<n;i++){
        cin>>u>>v;
        addedge(u,v);
        addedge(v,u);
    }
    dfs(1,0);
    for(int i=1;i<=n;i++){
        int t=n-sum[i];
        ans[i]+=1LL*t*(t+1)/2;
    }
    for(int i=1;i<=n;i++) cout<<1LL*n*(n+1)/2-ans[i]<<'\n';
}
void dfs(int u,int fa){
    siz[u]=1;
    int re=0;
    for(int i=he[u];i;i=nex[i]){
        int v=ver[i];
        if(v==fa) continue;
        int last=sum[c[u]];
        dfs(v,u);
        siz[u]+=siz[v];
        int t=sum[c[u]]-last;
        ans[c[u]]+=1LL*(siz[v]-t)*(siz[v]-t+1)/2;
        re+=siz[v]-t;
    }
    sum[c[u]]+=re+1;
}
void addedge(int u,int v){
    ver[tot]=v;nex[tot]=he[u];he[u]=tot++;
}

### 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<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;

const int N=2e5+5;
int he[N],ver[N*2],nex[N*2],tot;
void addedge(int,int);
int c[N];
int sum[N],siz[N];
ll ans[N];
void dfs(int,int);

void init();
void work();
int ca=1;
int main(){
//    freopen("in.txt","r",stdin);
    ios::sync_with_stdio(0);
    cin.tie(0),cout.tie(0);
//    init();
//    int T;cin>>T;
//    while(T--)
    work();
    return 0;
}
void work(){
    int n;cin>>n;
    for(int i=1;i<=n;i++) cin>>c[i];
    int u,v;tot=1;
    for(int i=1;i<n;i++){
        cin>>u>>v;
        addedge(u,v);
        addedge(v,u);
    }
    dfs(1,0);
    for(int i=1;i<=n;i++){
        int t=n-sum[i];
        ans[i]+=1LL*t*(t+1)/2;
    }
    for(int i=1;i<=n;i++) cout<<1LL*n*(n+1)/2-ans[i]<<'\n';
}
void dfs(int u,int fa){
    siz[u]=1;
    int re=0;
    for(int i=he[u];i;i=nex[i]){
        int v=ver[i];
        if(v==fa) continue;
        int last=sum[c[u]];
        dfs(v,u);
        siz[u]+=siz[v];
        int t=sum[c[u]]-last;
        ans[c[u]]+=1LL*(siz[v]-t)*(siz[v]-t+1)/2;
        re+=siz[v]-t;
    }
    sum[c[u]]+=re+1;
}
void addedge(int u,int v){
    ver[tot]=v;nex[tot]=he[u];he[u]=tot++;
}

```