#include<bits/stdc++.h>
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
using namespace std;
typedef long long ll;
const int N=2e5+10;
vector<int>G[N];
int n,col[N],cnt[N],all;
ll ans[N],sz[N];
ll dfs(int u,int fa)
{
    ll tmp=cnt[col[u]];//保存其他树的cnt
    sz[u]=1;
    for(int v:G[u]){
        if(v==fa) continue;
        ll l=all-cnt[col[u]];//子树前
        sz[u]+=dfs(v,u);
        ll r=all-cnt[col[u]];//遍历子树后
        ll len=r-l;
        ans[col[u]]-=(len*(len-1)/2);
    }
    cnt[col[u]]=tmp+sz[u];//每次col[u]的颜色  重新被当前u全部覆盖
    all++;
    return sz[u];
}
int main()
{
    scanf("%d",&n);
    rep(i,1,n) scanf("%d",&col[i]);
    rep(i,2,n)
    {
        int u,v;scanf("%d%d",&u,&v);
        G[u].push_back(v);
        G[v].push_back(u);
    }
 
 
 
    rep(i,1,n) ans[i]=1ll*n*(n-1)/2;//所有路径
    dfs(1,-1);
    for(int i=1;i<=n;++i){//除了子树，从根节点到最近的颜色 联通块也要去掉
        //printf("i:%d cnt:%d ans:%lld\n",i,cnt[i],ans[i]);
        ll len=n-cnt[i];
        ans[i]-=(len*(len-1)/2);
    }
    for(int i=1;i<=n;++i) ans[col[i]]++;//每个单独的点是一条路径
    for(int i=1;i<=n;++i) printf("%lld\n",ans[i]);
}
### 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>
#define rep(i,a,b) for(int i=(a);i<=(b);++i)
using namespace std;
typedef long long ll;
const int N=2e5+10;
vector<int>G[N];
int n,col[N],cnt[N],all;
ll ans[N],sz[N];
ll dfs(int u,int fa)
{
    ll tmp=cnt[col[u]];//保存其他树的cnt
    sz[u]=1;
    for(int v:G[u]){
        if(v==fa) continue;
        ll l=all-cnt[col[u]];//子树前
        sz[u]+=dfs(v,u);
        ll r=all-cnt[col[u]];//遍历子树后
        ll len=r-l;
        ans[col[u]]-=(len*(len-1)/2);
    }
    cnt[col[u]]=tmp+sz[u];//每次col[u]的颜色  重新被当前u全部覆盖
    all++;
    return sz[u];
}
int main()
{
    scanf("%d",&n);
    rep(i,1,n) scanf("%d",&col[i]);
    rep(i,2,n)
    {
        int u,v;scanf("%d%d",&u,&v);
        G[u].push_back(v);
        G[v].push_back(u);
    }
 
 
 
    rep(i,1,n) ans[i]=1ll*n*(n-1)/2;//所有路径
    dfs(1,-1);
    for(int i=1;i<=n;++i){//除了子树，从根节点到最近的颜色 联通块也要去掉
        //printf("i:%d cnt:%d ans:%lld\n",i,cnt[i],ans[i]);
        ll len=n-cnt[i];
        ans[i]-=(len*(len-1)/2);
    }
    for(int i=1;i<=n;++i) ans[col[i]]++;//每个单独的点是一条路径
    for(int i=1;i<=n;++i) printf("%lld\n",ans[i]);
}
```