#include<bits/stdc++.h>
#define ll long long
#define N 200005
#define M 1000000007
#define pii pair<ll,ll>
#define fast ios_base::sync_with_stdio(0);cin.tie(0);
using namespace std;
vector<ll>e[N],all[N];
map<ll,ll>mp;
ll n;
ll pr[N];
ll c[N];
ll level[N];
ll sub[N];
bool cmp(ll a,ll b)
{
    return level[a]>level[b];
}

void dfs(ll node,ll parent,ll lv=1)
{
    ll gone=mp[c[node]];
    ll now=0;
    ll ans=0;
    sub[node]=1;
    ll last=gone;
   // cout<<node<<' '<<gone<<endl;
    for(ll u:e[node])
    {
        if(u!=parent)
        {
            dfs(u,node);
            ll extra=mp[c[node]]-last;
            ans+=now*(sub[u]-extra);
            now+=sub[u]-extra;
            sub[node]+=sub[u];
            last=mp[c[node]];
        }
    }
    //cout<<node<<' '<<ans<<endl;
    ll upore=n-gone-sub[node]+1;
    ll niche=sub[node]-(mp[c[node]]-gone);
    ans+=upore*niche;
    //cout<<node<<' '<<upore*niche<<' '<<ans<<endl;
    pr[c[node]]+=ans;
    mp[c[node]]=gone+sub[node];
    return;
}
int main()
{
    cin>>n;
    for(ll i=1;i<=n;i++)
        cin>>c[i];
    for(ll i=1;i<=n-1;i++)
    {
        ll a,b;
        cin>>a>>b;
        e[a].push_back(b);
        e[b].push_back(a);
    }
    dfs(1,0);
    for(ll i=1;i<=n;i++)
        cout<<pr[i]<<'\n';
}

### Prompt

Please create a solution in CPP 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 ll long long
#define N 200005
#define M 1000000007
#define pii pair<ll,ll>
#define fast ios_base::sync_with_stdio(0);cin.tie(0);
using namespace std;
vector<ll>e[N],all[N];
map<ll,ll>mp;
ll n;
ll pr[N];
ll c[N];
ll level[N];
ll sub[N];
bool cmp(ll a,ll b)
{
    return level[a]>level[b];
}

void dfs(ll node,ll parent,ll lv=1)
{
    ll gone=mp[c[node]];
    ll now=0;
    ll ans=0;
    sub[node]=1;
    ll last=gone;
   // cout<<node<<' '<<gone<<endl;
    for(ll u:e[node])
    {
        if(u!=parent)
        {
            dfs(u,node);
            ll extra=mp[c[node]]-last;
            ans+=now*(sub[u]-extra);
            now+=sub[u]-extra;
            sub[node]+=sub[u];
            last=mp[c[node]];
        }
    }
    //cout<<node<<' '<<ans<<endl;
    ll upore=n-gone-sub[node]+1;
    ll niche=sub[node]-(mp[c[node]]-gone);
    ans+=upore*niche;
    //cout<<node<<' '<<upore*niche<<' '<<ans<<endl;
    pr[c[node]]+=ans;
    mp[c[node]]=gone+sub[node];
    return;
}
int main()
{
    cin>>n;
    for(ll i=1;i<=n;i++)
        cin>>c[i];
    for(ll i=1;i<=n-1;i++)
    {
        ll a,b;
        cin>>a>>b;
        e[a].push_back(b);
        e[b].push_back(a);
    }
    dfs(1,0);
    for(ll i=1;i<=n;i++)
        cout<<pr[i]<<'\n';
}

```