#include<bits/stdc++.h>
#define see(x) cerr<<#x<<"="<<x<<endl;
using namespace std;
typedef long long ll;
const ll mod=1e9+7;
ll fast(ll x,ll n) {ll ans=1;for(;n;n>>=1,x=x*x%mod) if(n&1) ans=ans*x%mod; return ans;}
const int N=1e6+100;
inline int read() {int x=0,f=1;char c=getchar();while(c!='-'&&(c<'0'||c>'9'))c=getchar();if(c=='-')f=-1,c=getchar();while(c>='0'&&c<='9')x=x*10+c-'0',c=getchar();return f*x;}
ll c(ll x) {
    return x*(x+1)/2;
}
ll ans[N],siz[N],temp[N],a[N],n;
vector<int>edge[N];

void dfs(int x,int f) {
    ll t=temp[a[x]];
    siz[x]=1;
    for(auto v:edge[x]) if(v!=f) {
        ll pre=temp[a[x]];
        dfs(v,x);siz[x]+=siz[v];
        ans[a[x]]-=c(siz[v]-(temp[a[x]]-pre));
    }
    temp[a[x]]=t+siz[x];
}
int main() {
    cin>>n;
    for(int i=1;i<=n;i++) a[i]=read();
    for(int i=1;i<n;i++) {
        int u,v;
        u=read();v=read();
        //scanf("%d%d",&u,&v);
        edge[u].push_back(v);
        edge[v].push_back(u);
    }
    dfs(1,0);
    for(int i=1;i<=n;i++) {
        ans[i]+=c(n)-c(n-temp[i]);
        printf("%lld\n",ans[i]);
    }
}

### 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>
#define see(x) cerr<<#x<<"="<<x<<endl;
using namespace std;
typedef long long ll;
const ll mod=1e9+7;
ll fast(ll x,ll n) {ll ans=1;for(;n;n>>=1,x=x*x%mod) if(n&1) ans=ans*x%mod; return ans;}
const int N=1e6+100;
inline int read() {int x=0,f=1;char c=getchar();while(c!='-'&&(c<'0'||c>'9'))c=getchar();if(c=='-')f=-1,c=getchar();while(c>='0'&&c<='9')x=x*10+c-'0',c=getchar();return f*x;}
ll c(ll x) {
    return x*(x+1)/2;
}
ll ans[N],siz[N],temp[N],a[N],n;
vector<int>edge[N];

void dfs(int x,int f) {
    ll t=temp[a[x]];
    siz[x]=1;
    for(auto v:edge[x]) if(v!=f) {
        ll pre=temp[a[x]];
        dfs(v,x);siz[x]+=siz[v];
        ans[a[x]]-=c(siz[v]-(temp[a[x]]-pre));
    }
    temp[a[x]]=t+siz[x];
}
int main() {
    cin>>n;
    for(int i=1;i<=n;i++) a[i]=read();
    for(int i=1;i<n;i++) {
        int u,v;
        u=read();v=read();
        //scanf("%d%d",&u,&v);
        edge[u].push_back(v);
        edge[v].push_back(u);
    }
    dfs(1,0);
    for(int i=1;i<=n;i++) {
        ans[i]+=c(n)-c(n-temp[i]);
        printf("%lld\n",ans[i]);
    }
}

```