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 n, a[1005000];
vector<int> e[1005000];
map<int, int> cnt[1005000];
map<int, long long> sum[1005000], ans;
void dfs(int x, int p) {
  cnt[x][a[x]] = 1;
  sum[x][1] = a[x];
  for (int y : e[x]) {
    if (y == p) continue;
    dfs(y, x);
    int fi = x;
    int se = y;
    if (cnt[fi].size() < cnt[se].size()) swap(fi, se);
    for (auto i : cnt[se]) {
      sum[fi][cnt[fi][i.first]] -= i.first;
      cnt[fi][i.first] += i.second;
      sum[fi][cnt[fi][i.first]] += i.first;
    }
    cnt[se].clear();
    sum[se].clear();
    if (x != fi) {
      cnt[x].swap(cnt[fi]);
      sum[x].swap(sum[fi]);
    }
  }
  ans[x] = sum[x].rbegin()->second;
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> a[i];
  for (int i = 1; i < n; i++) {
    int x, y;
    cin >> x >> y;
    e[x].push_back(y);
    e[y].push_back(x);
  }
  dfs(1, 0);
  for (int i = 1; i <= n; i++) cout << ans[i] << ' ';
  return 0;
}
