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 int N = 1e5 + 5;
int n, a, b, cnt = 0;
vector<int> color(N), pos(N);
vector<long long> ans(N);
vector<map<long long, long long> > sets(N), Cnt(N);
vector<vector<int> > adj(N);
void dfs(int par, int u) {
  int children = 0, mx = 0, ch;
  for (int v : adj[u]) {
    if (v != par) {
      ++children;
      dfs(u, v);
      if (sets[pos[v]].size() > mx) {
        mx = sets[pos[v]].size();
        ch = v;
      }
    }
  }
  if (!children) {
    ++cnt;
    ++sets[cnt][color[u]];
    Cnt[cnt][sets[cnt][color[u]]] += color[u];
    pos[u] = cnt;
  } else {
    for (int v : adj[u]) {
      if (v != par && v != ch) {
        for (auto& p : sets[pos[v]]) {
          sets[pos[ch]][p.first] += p.second;
          Cnt[pos[ch]][sets[pos[ch]][p.first]] += p.first;
        }
        sets[pos[v]].clear();
        Cnt[pos[v]].clear();
      }
    }
    ++sets[pos[ch]][color[u]];
    Cnt[pos[ch]][sets[pos[ch]][color[u]]] += color[u];
    pos[u] = pos[ch];
  }
  ans[u] = Cnt[pos[u]].rbegin()->second;
}
int main(void) {
  ios ::sync_with_stdio(0);
  cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) {
    cin >> color[i];
  }
  for (int i = 1; i <= n - 1; ++i) {
    cin >> a >> b;
    adj[a].emplace_back(b);
    adj[b].emplace_back(a);
  }
  dfs(0, 1);
  for (int i = 1; i <= n; ++i) {
    cout << ans[i] << " \n"[i == n];
  }
  return 0;
}
