p02710 AtCoder Beginner Contest 163 - path pass i_33241
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
#include<iostream>
#include<algorithm>
#include<vector>
#include<cstdint>

int const MAXN = 2e5;
int c[MAXN];
std::vector<int> G[MAXN];
int N;
int64_t ans[MAXN];
int nn[MAXN], csum[MAXN];

void dfs(int v, int p) {
  int cv = c[v], save = csum[cv], nodes = 1;
  for(auto ch: G[v]) if(ch != p) {
    int pcsum = csum[cv];
    dfs(ch, v);
    int dcsum = csum[cv] - pcsum;
    int w = nn[ch] - dcsum;
    ans[cv] -= (int64_t)w*(w+1)/2;
    nodes += nn[ch];
  }
  csum[cv] = save + (nn[v] = nodes);
}

int main() {
  std::cin >> N;
  for(int i = 0; i < N; ++i) { std::cin >> c[i]; --c[i]; }
  for(int i = 0; i < N-1; ++i) {
    int a, b; std::cin >> a >> b; --a; --b;
    G[a].push_back(b);
    G[b].push_back(a);
  }
  
  std::fill_n(ans, N, (int64_t)N*(N+1)/2);
  dfs(0, -1);
  for(int i = 0; i < N; ++i) {
    int w = N - csum[i];
    ans[i] -= (int64_t)w*(w+1)/2;
  }
  for(int i = 0; i < N; ++i) std::cout << ans[i] << std::endl;
  return 0;
}
