#include <bits/stdc++.h>

using namespace std;
typedef long long ll;

const int N=200005;
ll n;
vector<int> v[N];
vector<int> c(N,0);
vector<int> sz(N,0);
vector<int> big(N,0);
vector<ll> res(N,0);

void dfs(int a,int p=0)
{
    sz[a]=1;
    for(int to:v[a])
    {
        if(to==p) continue;
        dfs(to,a);
        sz[a]+=sz[to];
        if(sz[to]>sz[big[a]]) big[a]=to;
    }
}

map<int,ll> solve(int a,int p=0)
{
    map<int,ll> m;
    if(big[a]!=0) m=solve(big[a],a);
    for(int to:v[a])
    {
        if(to==p||to==big[a]) continue;
        map<int,ll> x=solve(to,a);
        for(auto [col,cnt]:x) m[col]+=cnt;
    }
    m[c[a]]=sz[a];
    if(p!=0)
    {
        ll x=sz[a];
        if(m.find(c[p])!=m.end()) x-=m[c[p]];
        res[c[p]]-=(x*(x-1)/2+x);
    }
    else
    {
        for(int i=1;i<=n;i++)
        {
            ll x=sz[a]-m[i];
            res[i]-=(x*(x-1)/2+x);
        }
    }
    return m;
}

int main()
{
    scanf("%lld",&n);
    for(int i=0;i<N;i++) res[i]=(n*(n-1)/2+n);
    for(int i=1;i<=n;i++) scanf("%d",&c[i]);
    for(int i=0;i<n-1;i++)
    {
        int a,b;
        scanf("%d%d",&a,&b);
        v[a].push_back(b);
        v[b].push_back(a);
    }
    dfs(1);
    solve(1);
    for(int i=1;i<=n;i++) printf("%lld\n",res[i]);
    return 0;
}

### Prompt

Construct a cpp code solution to the problem outlined:

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=200005;
ll n;
vector<int> v[N];
vector<int> c(N,0);
vector<int> sz(N,0);
vector<int> big(N,0);
vector<ll> res(N,0);

void dfs(int a,int p=0)
{
    sz[a]=1;
    for(int to:v[a])
    {
        if(to==p) continue;
        dfs(to,a);
        sz[a]+=sz[to];
        if(sz[to]>sz[big[a]]) big[a]=to;
    }
}

map<int,ll> solve(int a,int p=0)
{
    map<int,ll> m;
    if(big[a]!=0) m=solve(big[a],a);
    for(int to:v[a])
    {
        if(to==p||to==big[a]) continue;
        map<int,ll> x=solve(to,a);
        for(auto [col,cnt]:x) m[col]+=cnt;
    }
    m[c[a]]=sz[a];
    if(p!=0)
    {
        ll x=sz[a];
        if(m.find(c[p])!=m.end()) x-=m[c[p]];
        res[c[p]]-=(x*(x-1)/2+x);
    }
    else
    {
        for(int i=1;i<=n;i++)
        {
            ll x=sz[a]-m[i];
            res[i]-=(x*(x-1)/2+x);
        }
    }
    return m;
}

int main()
{
    scanf("%lld",&n);
    for(int i=0;i<N;i++) res[i]=(n*(n-1)/2+n);
    for(int i=1;i<=n;i++) scanf("%d",&c[i]);
    for(int i=0;i<n-1;i++)
    {
        int a,b;
        scanf("%d%d",&a,&b);
        v[a].push_back(b);
        v[b].push_back(a);
    }
    dfs(1);
    solve(1);
    for(int i=1;i<=n;i++) printf("%lld\n",res[i]);
    return 0;
}

```