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;
long long q, a[555555], par[555555], f[555555], cnt[555555], ans[555555], k, l,
    m, n, o, p;
map<long long, long long> mp[555555];
vector<long long> adj[555555];
const long long mod = 1e9 + 7;
long long getPar(long long s) {
  if (s == par[s])
    return s;
  else
    return par[s] = getPar(par[s]);
}
void dfs(long long v) {
  mp[v][a[v]]++;
  ans[v] = a[v];
  cnt[v] = 1;
  for (auto u : adj[v]) {
    if (u == f[v]) continue;
    f[u] = v;
    dfs(u);
    o = getPar(v);
    p = getPar(u);
    if (mp[o].size() < mp[p].size()) {
      swap(o, p);
      cnt[v] = cnt[u];
      ans[v] = ans[u];
    }
    par[p] = o;
    for (auto op : mp[p]) {
      mp[o][op.first] += op.second;
      if (mp[o][op.first] > cnt[v]) {
        cnt[v] = mp[o][op.first];
        ans[v] = op.first;
      } else if (mp[o][op.first] == cnt[v]) {
        ans[v] += op.first;
      }
    }
  }
}
void solve() {
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    par[i] = i;
    cin >> a[i];
  }
  for (long long i = 1; i < n; i++) {
    cin >> o >> p;
    adj[o].push_back(p);
    adj[p].push_back(o);
  }
  dfs(1);
  for (long long i = 1; i <= n; i++) {
    cout << ans[i] << " ";
  }
}
int main() {
  ios_base::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  q = 1;
  while (q--) {
    solve();
  }
}
