#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int N=2e5+5;
int n,tot,si[N],fa[N],f[N],head[N],nex[N<<1],to[N<<1],a[N];
void add(int u,int v){to[++tot]=v;nex[tot]=head[u];head[u]=tot;}
vector<int>v[N];
ll ans[N],s[N];
int top,st[N];
void dfs1(int u,int p)
{
    si[u]=1;fa[u]=p;
    for(int i=head[u];i;i=nex[i])
    {
        int v=to[i];if(v==p)continue;
        dfs1(v,u);
        si[u]+=si[v];
    }
}
void dfs2(int u,int p)
{
    if(u==1||a[u]==a[p]) f[u]=0;
    else
    {
        f[u]=si[u];
        if(v[a[u]].size())
        {
            f[st[v[a[u]].back()+1]]-=si[u];
        }
    }
    st[++top]=u;
    v[a[u]].push_back(top);
    for(int i=head[u];i;i=nex[i])
    {
        int v=to[i];if(v==p)continue;
        dfs2(v,u);
    }
    top--;
    v[a[u]].pop_back();
}
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]),s[a[i]]--,s[i]+=n;
    for(int i=1;i<n;i++)
    {
        int u,v;scanf("%d%d",&u,&v);add(u,v);add(v,u);
    }
    dfs1(1,0);
    dfs2(1,0);
    for(int i=2;i<=n;i++)
    {
        if(f[i]==0) continue;
        int c=a[fa[i]];
        s[c]-=f[i];
        ans[c]+=(ll)f[i]*(f[i]-1)/2+f[i];
    }
    for(int i=1;i<=n;i++)
    {
        ans[i]+=(ll)s[i]*(s[i]-1)/2+s[i];
        printf("%lld\n",(ll)n*(n-1)/2+n-ans[i]);
    }
}

### 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;
typedef long long ll;
const int N=2e5+5;
int n,tot,si[N],fa[N],f[N],head[N],nex[N<<1],to[N<<1],a[N];
void add(int u,int v){to[++tot]=v;nex[tot]=head[u];head[u]=tot;}
vector<int>v[N];
ll ans[N],s[N];
int top,st[N];
void dfs1(int u,int p)
{
    si[u]=1;fa[u]=p;
    for(int i=head[u];i;i=nex[i])
    {
        int v=to[i];if(v==p)continue;
        dfs1(v,u);
        si[u]+=si[v];
    }
}
void dfs2(int u,int p)
{
    if(u==1||a[u]==a[p]) f[u]=0;
    else
    {
        f[u]=si[u];
        if(v[a[u]].size())
        {
            f[st[v[a[u]].back()+1]]-=si[u];
        }
    }
    st[++top]=u;
    v[a[u]].push_back(top);
    for(int i=head[u];i;i=nex[i])
    {
        int v=to[i];if(v==p)continue;
        dfs2(v,u);
    }
    top--;
    v[a[u]].pop_back();
}
int main()
{
    scanf("%d",&n);
    for(int i=1;i<=n;i++)
        scanf("%d",&a[i]),s[a[i]]--,s[i]+=n;
    for(int i=1;i<n;i++)
    {
        int u,v;scanf("%d%d",&u,&v);add(u,v);add(v,u);
    }
    dfs1(1,0);
    dfs2(1,0);
    for(int i=2;i<=n;i++)
    {
        if(f[i]==0) continue;
        int c=a[fa[i]];
        s[c]-=f[i];
        ans[c]+=(ll)f[i]*(f[i]-1)/2+f[i];
    }
    for(int i=1;i<=n;i++)
    {
        ans[i]+=(ll)s[i]*(s[i]-1)/2+s[i];
        printf("%lld\n",(ll)n*(n-1)/2+n-ans[i]);
    }
}

```