You are given a rooted tree with root in vertex 1. Each vertex is coloured in some colour.

Let's call colour c dominating in the subtree of vertex v if there are no other colours that appear in the subtree of vertex v more times than colour c. So it's possible that two or more colours will be dominating in the subtree of some vertex.

The subtree of vertex v is the vertex v and all other vertices that contains vertex v in each path to the root.

For each vertex v find the sum of all dominating colours in the subtree of vertex v.

Input

The first line contains integer n (1 ≤ n ≤ 105) — the number of vertices in the tree.

The second line contains n integers ci (1 ≤ ci ≤ n), ci — the colour of the i-th vertex.

Each of the next n - 1 lines contains two integers xj, yj (1 ≤ xj, yj ≤ n) — the edge of the tree. The first vertex is the root of the tree.

Output

Print n integers — the sums of dominating colours for each vertex.

Examples

Input

4
1 2 3 4
1 2
2 3
2 4


Output

10 9 3 4


Input

15
1 2 3 1 2 3 3 1 1 3 2 2 1 2 3
1 2
1 3
1 4
1 14
1 15
2 5
2 6
2 7
3 8
3 9
3 10
4 11
4 12
4 13


Output

6 5 4 3 2 3 3 1 1 3 2 2 1 2 3
#include <bits/stdc++.h>
using namespace std;
int dx[4] = {-1, 1, 0, 0}, dy[4] = {0, 0, -1, 1};
vector<long long int> ans(100002), mx(100002), col(100002), vec[100002];
map<int, int> mp[100002];
void solve(int node, int ata) {
  for (auto it : vec[node]) {
    if (it != ata) {
      solve(it, node);
      if (mp[it].size() > mp[node].size()) {
        swap(mp[it], mp[node]);
        ans[node] = ans[it];
        mx[node] = mx[it];
      }
      for (auto p : mp[it]) {
        mp[node][p.first] += p.second;
        if (mp[node][p.first] == mx[node]) {
          ans[node] += p.first;
        } else if (mp[node][p.first] > mx[node]) {
          mx[node] = mp[node][p.first];
          ans[node] = p.first;
        }
      }
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  int n;
  cin >> n;
  for (int i = 1; n >= i; i++) {
    cin >> col[i];
    mp[i][col[i]] = 1;
    ans[i] = col[i];
    mx[i] = 1;
  }
  for (int i = 1; n > i; i++) {
    int a, b;
    cin >> a >> b;
    vec[a].push_back(b);
    vec[b].push_back(a);
  }
  solve(1, 0);
  for (int i = 1; n >= i; i++) {
    cout << ans[i] << " ";
  }
}
