#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 2e5+5;
const int mod = 1e9+7;
ll qpow(ll a,ll b){ll res=1;for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
struct graph
{
    int head[maxn],nxt[maxn<<1],to[maxn<<1],w[maxn<<1],sz;
    void init(){memset(head,-1,sizeof(head));}
    graph(){init();}
    void push(int a,int b,int c){nxt[sz]=head[a],to[sz]=b,w[sz]=c,head[a]=sz++;}
    int& operator[](const int a){return to[a];}
}g;
int a[maxn],sum[maxn],SIZE[maxn];
ll ans[maxn];
ll get(int x)
{
    return 1ll*x*(x+1)/2;
}
void dfs(int now,int pre)
{
    SIZE[now]=1;
    int tmp = sum[a[now]];
    for(int i = g.head[now];~i;i = g.nxt[i]){
        if(g[i]==pre)continue;
        sum[a[now]]=0;
        dfs(g[i],now);
        SIZE[now]+=SIZE[g[i]];
        ans[a[now]]-=get(SIZE[g[i]]-sum[a[now]]);
    }
    sum[a[now]]=tmp+SIZE[now];
}
int main()
{
    int n;
    scanf("%d",&n);
    for(int i = 1;i <= n;++i)scanf("%d",&a[i]);
    for(int i = 1,a,b;i < n;++i){
        scanf("%d%d",&a,&b);
        g.push(a,b,0);
        g.push(b,a,0);
    }
    for(int i = 1;i <= n;++i)ans[i]=get(n);
    dfs(1,0);
    for(int i = 1;i <= n;++i){
        ans[i]-=get(n-sum[i]);
    }
    for(int i = 1;i <= n;++i)printf("%lld\n",ans[i]);
    return 0;
}

### Prompt

Generate 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>
using namespace std;
typedef long long ll;
const int maxn = 2e5+5;
const int mod = 1e9+7;
ll qpow(ll a,ll b){ll res=1;for(;b;b>>=1){if(b&1)res=res*a%mod;a=a*a%mod;}return res;}
struct graph
{
    int head[maxn],nxt[maxn<<1],to[maxn<<1],w[maxn<<1],sz;
    void init(){memset(head,-1,sizeof(head));}
    graph(){init();}
    void push(int a,int b,int c){nxt[sz]=head[a],to[sz]=b,w[sz]=c,head[a]=sz++;}
    int& operator[](const int a){return to[a];}
}g;
int a[maxn],sum[maxn],SIZE[maxn];
ll ans[maxn];
ll get(int x)
{
    return 1ll*x*(x+1)/2;
}
void dfs(int now,int pre)
{
    SIZE[now]=1;
    int tmp = sum[a[now]];
    for(int i = g.head[now];~i;i = g.nxt[i]){
        if(g[i]==pre)continue;
        sum[a[now]]=0;
        dfs(g[i],now);
        SIZE[now]+=SIZE[g[i]];
        ans[a[now]]-=get(SIZE[g[i]]-sum[a[now]]);
    }
    sum[a[now]]=tmp+SIZE[now];
}
int main()
{
    int n;
    scanf("%d",&n);
    for(int i = 1;i <= n;++i)scanf("%d",&a[i]);
    for(int i = 1,a,b;i < n;++i){
        scanf("%d%d",&a,&b);
        g.push(a,b,0);
        g.push(b,a,0);
    }
    for(int i = 1;i <= n;++i)ans[i]=get(n);
    dfs(1,0);
    for(int i = 1;i <= n;++i){
        ans[i]-=get(n-sum[i]);
    }
    for(int i = 1;i <= n;++i)printf("%lld\n",ans[i]);
    return 0;
}

```