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;
const long long int M = 1000000007;
vector<int> adj[100001];
map<int, int> mp[100001];
int arr[100001];
long long int ans[100001];
long long int mx[100001];
int n;
void dfs(int cur, int par) {
  mp[cur][arr[cur]]++;
  for (auto first : adj[cur])
    if (first != par) {
      dfs(first, cur);
      if (mp[cur].size() < mp[first].size()) {
        swap(mp[cur], mp[first]);
        ans[cur] = ans[first];
        mx[cur] = mx[first];
      }
      for (auto n : mp[first]) {
        mp[cur][n.first] += n.second;
        if (mx[cur] < mp[cur][n.first] and n.first != arr[cur]) {
          ans[cur] = n.first;
          mx[cur] = mp[cur][n.first];
        } else if (mx[cur] == mp[cur][n.first] and n.first != arr[cur])
          ans[cur] += n.first;
      }
      mp[first].clear();
    }
  if (mx[cur] < mp[cur][arr[cur]]) {
    ans[cur] = arr[cur];
    mx[cur] = mp[cur][arr[cur]];
  } else if (mx[cur] == mp[cur][arr[cur]])
    ans[cur] += arr[cur];
  if (mx[cur] == -1) ans[cur] = arr[cur];
}
void solve() {
  for (int i = 0; i < 100001; i++) mx[i] = -1;
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> arr[i];
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs(1, 0);
  for (int i = 1; i <= n; i++) cout << ans[i] << " ";
  cout << endl;
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  long long int t = 1;
  while (t--) solve();
}
