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 N = 1e5 + 5;
long long n;
vector<long long> sz(N, 0);
vector<long long> adj[N];
void getsize(long long node, long long par) {
  sz[node] = 1;
  for (auto j : adj[node]) {
    if (j != par) {
      getsize(j, node);
      sz[node] += sz[j];
    }
  }
}
vector<long long> col(N, 0);
map<long long, long long> answer;
vector<long long> cnt(N, 0);
vector<long long> v[N];
vector<long long> answer_for_node(N, 0);
void dfs(long long node, long long par, long long keep) {
  long long maxsize = 0, bigchild = -1;
  for (auto j : adj[node]) {
    if (j != par) {
      if (maxsize < sz[j]) {
        maxsize = sz[j];
        bigchild = j;
      }
    }
  }
  for (auto j : adj[node]) {
    if (j != par and j != bigchild) {
      dfs(j, node, 0);
    }
  }
  if (bigchild != -1) {
    dfs(bigchild, node, 1);
    swap(v[node], v[bigchild]);
  }
  v[node].push_back(node);
  answer[cnt[col[node]]] -= col[node];
  cnt[col[node]]++;
  answer[cnt[col[node]]] += col[node];
  for (auto j : adj[node]) {
    if (j != par and j != bigchild) {
      for (auto k : v[j]) {
        v[node].push_back(k);
        answer[cnt[col[k]]] -= col[k];
        cnt[col[k]]++;
        answer[cnt[col[k]]] += col[k];
      }
    }
  }
  answer_for_node[node] = answer.rbegin()->second;
  if (keep == 0) {
    for (auto j : v[node]) {
      answer[cnt[col[j]]] -= col[j];
      if (answer[cnt[col[j]]] == 0) {
        answer.erase(cnt[col[j]]);
      }
      cnt[col[j]]--;
      answer[cnt[col[j]]] += col[j];
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(nullptr);
  cin >> n;
  for (long long i = 0; i < n; ++i) {
    cin >> col[i];
  }
  for (long long i = 0; i < n - 1; ++i) {
    long long a, b;
    cin >> a >> b;
    a--;
    b--;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  getsize(0, -1);
  dfs(0, -1, 0);
  for (long long i = 0; i < n; ++i) {
    cout << answer_for_node[i] << " ";
  }
  cout << '\n';
  return 0;
}
