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 + 1;
map<int, int> m[N];
set<pair<int, int> > s[N];
vector<int> g[N];
int n, c[N];
long long ans[N];
void dfs(int u, int p) {
  m[u][c[u]] = 1;
  s[u].insert(pair<int, int>(1, c[u]));
  ans[u] = c[u];
  for (int v : g[u]) {
    if (v == p) continue;
    dfs(v, u);
    if (m[u].size() < m[v].size()) {
      m[u].swap(m[v]);
      s[u].swap(s[v]);
      ans[u] = ans[v];
    }
  }
  for (int v : g[u]) {
    if (v == p) continue;
    for (auto x : m[v]) {
      s[u].erase(pair<int, int>(m[u][x.first], x.first));
      m[u][x.first] += x.second;
      if (s[u].empty() || m[u][x.first] > (*--s[u].end()).first)
        ans[u] = x.first;
      else if (m[u][x.first] == (*--s[u].end()).first)
        ans[u] += x.first;
      s[u].insert(pair<int, int>(m[u][x.first], x.first));
    }
  }
}
int main() {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> c[i];
  for (int i = 1; i < n; ++i) {
    int x, y;
    cin >> x >> y;
    g[x].push_back(y);
    g[y].push_back(x);
  }
  dfs(1, 0);
  for (int i = 1; i <= n; ++i) cout << ans[i] << " ";
}
