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 = 1000 * 100;
int col[N + 3], sz[N + 3];
long long ans[N + 3];
map<int, long long>* cnt[N + 2];
map<int, long long>* val[N + 3];
vector<int> g[N + 3];
void dfs(int u, int p) {
  int mx = -1, bigChild = -1;
  sz[u] = 1;
  for (auto v : g[u]) {
    if (v == p) continue;
    dfs(v, u);
    sz[u] += sz[v];
    if (sz[v] > mx) mx = sz[v], bigChild = v;
  }
  if (bigChild != -1) {
    cnt[u] = cnt[bigChild], val[u] = val[bigChild];
  } else {
    cnt[u] = new map<int, long long>(), val[u] = new map<int, long long>();
  }
  (*val[u])[(*cnt[u])[col[u]]] -= col[u];
  (*cnt[u])[col[u]] += 1;
  (*val[u])[(*cnt[u])[col[u]]] += col[u];
  for (auto v : g[u]) {
    if (v == p || v == bigChild) continue;
    for (auto it : (*cnt[v])) {
      int colr = it.first;
      long long freq = it.second;
      (*val[u])[(*cnt[u])[colr]] -= colr;
      (*cnt[u])[colr] += freq;
      (*val[u])[(*cnt[u])[colr]] += colr;
    }
  }
  ans[u] = ((*val[u]).rbegin())->second;
  return;
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  memset(sz, 0, sizeof sz);
  int n;
  cin >> n;
  for (int i = 1; i <= n; ++i) cin >> col[i];
  for (int i = 1; i < n; ++i) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1, 0);
  for (int i = 1; i <= n; ++i) cout << ans[i] << " ";
  cout << "\n";
  return 0;
}
