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;
class Task {
 public:
  vector<int> color;
  vector<vector<int>> g;
  vector<long long> ans;
  class subtree {
   public:
    unordered_map<int, int> color_cnt;
    int max;
    long long max_sum;
  };
  subtree* merge(subtree* a, subtree* b) {
    if (((int)a->color_cnt.size()) < ((int)b->color_cnt.size())) {
      swap(a, b);
    }
    for (auto p : b->color_cnt) {
      int c = p.first;
      int c_cnt = p.second;
      a->color_cnt[c] += c_cnt;
      if (a->color_cnt[c] > a->max) {
        a->max = a->color_cnt[c];
        a->max_sum = c;
      } else if (a->color_cnt[c] == a->max) {
        a->max_sum += c;
      }
    }
    b->color_cnt.clear();
    return a;
  }
  subtree* dfs(int v, int p = -1) {
    subtree* res = new subtree();
    res->color_cnt[color[v]] = 1;
    res->max = 1;
    res->max_sum = color[v];
    for (int u : g[v]) {
      if (u == p) continue;
      res = merge(dfs(u, v), res);
    }
    ans[v] = res->max_sum;
    return res;
  }
  void solve() {
    int n;
    scanf("%d", &n);
    color.resize(n);
    for (int i = 0; i < n; i++) scanf("%d", &color[i]);
    g.resize(n);
    for (int i = 1; i < n; i++) {
      int u, v;
      scanf("%d %d", &u, &v);
      u--, v--;
      g[u].push_back(v);
      g[v].push_back(u);
    }
    ans.resize(n);
    dfs(0);
    for (int i = 0; i < n; i++) printf("%lld ", ans[i]);
    puts("");
  }
};
int main() {
  int T = 1;
  for (int t = 1; t <= T; t++) {
    (new Task())->solve();
  }
  return 0;
}
