#include<bits/stdc++.h>
#define pb push_back
#define ll long long
using namespace std;
void fastio() {ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);}
const int maxn = 2e5+5;
int col[maxn],sub[maxn],sz[maxn];
ll cnt[maxn];
vector<int> g[maxn];
int n,dfn;

void dfs(int u,int f)
{
    sz[u] = 1;
    int pre = sub[col[u]];
    for(int v:g[u])if(v!=f){
        int t = sub[col[u]];
        dfs(v,u);
        int dt = sub[col[u]] - t;
        int tmp = sz[v] - dt;
        cnt[col[u]] -= (ll)tmp*(tmp+1)/2;
        sz[u] += sz[v];
    }
    sub[col[u]] = pre+sz[u];
}

int main()
{
    fastio();
    cin>>n;
    for(int i=1;i<=n;i++) cnt[i] = (ll)n*(n+1)/2;
    for(int i=1;i<=n;i++) {cin>>col[i];}
    for(int i=1;i<n;i++){
        int u,v;
        cin>>u>>v;
        g[u].pb(v);g[v].pb(u);
    }
    dfs(1,-1);
    for(int i=1;i<=n;i++)
    {
        int t = n-sub[i];
        cnt[i] -= (ll) t*(t+1)/2;
        cout<<cnt[i]<<endl;
    }
    return 0;
}

### Prompt

Your challenge is to write 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 pb push_back
#define ll long long
using namespace std;
void fastio() {ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);}
const int maxn = 2e5+5;
int col[maxn],sub[maxn],sz[maxn];
ll cnt[maxn];
vector<int> g[maxn];
int n,dfn;

void dfs(int u,int f)
{
    sz[u] = 1;
    int pre = sub[col[u]];
    for(int v:g[u])if(v!=f){
        int t = sub[col[u]];
        dfs(v,u);
        int dt = sub[col[u]] - t;
        int tmp = sz[v] - dt;
        cnt[col[u]] -= (ll)tmp*(tmp+1)/2;
        sz[u] += sz[v];
    }
    sub[col[u]] = pre+sz[u];
}

int main()
{
    fastio();
    cin>>n;
    for(int i=1;i<=n;i++) cnt[i] = (ll)n*(n+1)/2;
    for(int i=1;i<=n;i++) {cin>>col[i];}
    for(int i=1;i<n;i++){
        int u,v;
        cin>>u>>v;
        g[u].pb(v);g[v].pb(u);
    }
    dfs(1,-1);
    for(int i=1;i<=n;i++)
    {
        int t = n-sub[i];
        cnt[i] -= (ll) t*(t+1)/2;
        cout<<cnt[i]<<endl;
    }
    return 0;
}

```