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 maxn = 1e5 + 100;
int n, cnt[maxn], cur_mx, sub[maxn], color[maxn];
long long amt[maxn], ans[maxn];
bool big[maxn];
vector<int> adj[maxn];
void dfsub(int v, int p = -1) {
  sub[v] = 1;
  for (auto to : adj[v]) {
    if (to == p) continue;
    dfsub(to, v);
    sub[v] += sub[to];
  }
}
void add(int v, int p, int delta) {
  amt[cnt[color[v]]] -= color[v];
  cnt[color[v]] += delta;
  amt[cnt[color[v]]] += color[v];
  if (delta == 1)
    cur_mx = max(cur_mx, cnt[color[v]]);
  else if (amt[cnt[color[v]] + 1] == 0)
    cur_mx--;
  for (auto to : adj[v]) {
    if (to == p || big[to]) continue;
    add(to, v, delta);
  }
}
void dfs(int v, int p, bool keep) {
  pair<int, int> mx = {-1, -1};
  for (auto to : adj[v])
    if (to != p) mx = max(mx, {sub[to], to});
  for (auto to : adj[v])
    if (to != p && to != mx.second) dfs(to, v, false);
  if (mx.second != -1) dfs(mx.second, v, true), big[mx.second] = true;
  add(v, p, 1);
  ans[v] = amt[cur_mx];
  if (mx.second != -1) big[mx.second] = false;
  if (!keep) add(v, p, -1);
}
int main() {
  ios_base::sync_with_stdio(false), cin.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> color[i];
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  dfsub(1);
  dfs(1, -1, 0);
  for (int i = 1; i <= n; i++) cout << ans[i] << " ";
  cout << "\n";
  return 0;
}
