#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
vector<int> g[200002];
LL ans[200002];
int sz[200002], c[200002];
int cnt[200002];
void dfs(int u, int p){
    sz[u] = 1;
    int tmp = cnt[c[u]];
    int sum = cnt[c[u]];
    for(int i=0;i<g[u].size();i++){
        int v = g[u][i];
        if(v == p) continue;
        dfs(v, u);
        sz[u] += sz[v];
        int co = cnt[c[u]]-tmp;
        ans[c[u]] -= (sz[v]-co)*1ll*(sz[v]-co-1)/2;
        tmp = cnt[c[u]];
    }
    cnt[c[u]] = sum+sz[u];
}
int main(){
    int n;
    scanf("%d", &n);
    for(int i=1;i<=n;i++) scanf("%d", &c[i]);
    for(int i=1;i<=n-1;i++){
        int a, b;
        scanf("%d%d", &a, &b);
        g[a].emplace_back(b);
        g[b].emplace_back(a);
    }
    for(int i=1;i<=n;i++) ans[i] = n*1ll*(n-1)/2;
    dfs(1, 0);
    for(int i=1;i<=n;i++) ans[i] -= (n-cnt[i])*1ll*(n-cnt[i]-1)/2;
    for(int i=1;i<=n;i++) ans[c[i]]++;
    for(int i=1;i<=n;i++) printf("%lld\n", ans[i]);
}
### Prompt

Create a solution in CPP for 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;
vector<int> g[200002];
LL ans[200002];
int sz[200002], c[200002];
int cnt[200002];
void dfs(int u, int p){
    sz[u] = 1;
    int tmp = cnt[c[u]];
    int sum = cnt[c[u]];
    for(int i=0;i<g[u].size();i++){
        int v = g[u][i];
        if(v == p) continue;
        dfs(v, u);
        sz[u] += sz[v];
        int co = cnt[c[u]]-tmp;
        ans[c[u]] -= (sz[v]-co)*1ll*(sz[v]-co-1)/2;
        tmp = cnt[c[u]];
    }
    cnt[c[u]] = sum+sz[u];
}
int main(){
    int n;
    scanf("%d", &n);
    for(int i=1;i<=n;i++) scanf("%d", &c[i]);
    for(int i=1;i<=n-1;i++){
        int a, b;
        scanf("%d%d", &a, &b);
        g[a].emplace_back(b);
        g[b].emplace_back(a);
    }
    for(int i=1;i<=n;i++) ans[i] = n*1ll*(n-1)/2;
    dfs(1, 0);
    for(int i=1;i<=n;i++) ans[i] -= (n-cnt[i])*1ll*(n-cnt[i]-1)/2;
    for(int i=1;i<=n;i++) ans[c[i]]++;
    for(int i=1;i<=n;i++) printf("%lld\n", ans[i]);
}
```