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 + 7;
long long C[N], S[N], Max[N], Color[N], cnt[N], St[N], En[N], id[N], Ans[N], t;
vector<long long> adj[N];
void dfs(long long u, long long p) {
  St[u] = ++t;
  id[t] = u;
  S[u] = 1;
  for (long long i = 0; i < adj[u].size(); i++) {
    long long v = adj[u][i];
    if (v == p) continue;
    dfs(v, u);
    S[u] += S[v];
    if (S[v] > S[adj[u][0]]) swap(adj[u][0], adj[u][i]);
  }
  En[u] = t;
}
void dfs(long long u, long long p, long long k) {
  for (long long i = 1; i < adj[u].size(); i++) {
    long long v = adj[u][i];
    if (v == p) continue;
    dfs(v, u, 0);
  }
  long long v = adj[u][0];
  if (v != p) dfs(v, u, 1);
  if (v != p) Max[u] = Max[v], Ans[u] = Ans[v];
  cnt[Color[u]]++;
  if (cnt[Color[u]] > Max[u])
    Max[u] = cnt[Color[u]], Ans[u] = Color[u];
  else if (cnt[Color[u]] == Max[u])
    Ans[u] += Color[u];
  for (long long i = 1; i < adj[u].size(); i++) {
    long long v = adj[u][i];
    if (v == p) continue;
    for (long long j = St[v]; j <= En[v]; j++) {
      long long w = id[j];
      cnt[Color[w]]++;
      if (cnt[Color[w]] > Max[u])
        Max[u] = cnt[Color[w]], Ans[u] = Color[w];
      else if (cnt[Color[w]] == Max[u])
        Ans[u] += Color[w];
    }
  }
  if (!k) {
    for (long long i = St[u]; i <= En[u]; i++) {
      cnt[Color[id[i]]]--;
    }
  }
}
int32_t main() {
  ios_base ::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  long long n;
  cin >> n;
  for (long long i = 1; i <= n; i++) {
    cin >> Color[i];
  }
  for (long long i = 1; i < n; i++) {
    long long u, v;
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  memset(Max, -1, sizeof Max);
  memset(Ans, -1, sizeof Ans);
  dfs(1, -1);
  dfs(1, -1, 1);
  for (long long i = 1; i <= n; i++) cout << Ans[i] << " ";
}
