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>
#pragma comment(linker, "/stack:247474112")
#pragma GCC optimize("Ofast")
using namespace std;
const long long maxn = 1e5 + 10;
unordered_map<long long, long long> *mp[maxn];
long long col[maxn], ans[maxn], count[maxn];
vector<long long> g[maxn];
long long mxtimes[maxn];
void dfs(long long aa, long long aaa = 0) {
  long long biggie = -1, mx = -1;
  for (long long a : g[aa]) {
    if (a == aaa) continue;
    dfs(a, aa);
    if ((mp[a]->size()) - mx > 0) {
      mx = mp[a]->size();
      biggie = a;
    }
    if (mxtimes[a] > mxtimes[aa]) {
      mxtimes[aa] = mxtimes[a];
      ans[aa] = ans[a];
    }
  }
  if (biggie == -1) {
    mxtimes[aa] = 1;
    ans[aa] = col[aa];
    mp[aa] = new unordered_map<long long, long long>();
    (*mp[aa])[col[aa]]++;
    return;
  } else {
    mp[aa] = mp[biggie];
    ans[aa] = ans[biggie];
    mxtimes[aa] = mxtimes[biggie];
    (*mp[aa])[col[aa]]++;
    if ((*mp[aa])[col[aa]] > mxtimes[aa]) {
      ans[aa] = col[aa];
      mxtimes[aa] = (*mp[aa])[col[aa]];
    } else if ((*mp[aa])[col[aa]] == mxtimes[aa]) {
      ans[aa] += col[aa];
    }
    for (long long a : g[aa]) {
      if (a == aaa or a == biggie) continue;
      for (auto first : *mp[a]) {
        (*mp[aa])[first.first] += first.second;
        if ((*mp[aa])[first.first] == mxtimes[aa]) {
          ans[aa] += first.first;
        } else if ((*mp[aa])[first.first] > mxtimes[aa]) {
          ans[aa] = first.first;
          mxtimes[aa] = ((*mp[aa])[first.first]);
        }
      }
    }
  }
}
int32_t main() {
  ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
  long long n;
  cin >> n;
  for (long long i = 1; i <= n; ++i) cin >> col[i];
  for (long long i = 1; i < n; ++i) {
    long long u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1);
  for (long long i = 1; i <= n; ++i) cout << ans[i] << ' ';
  return 0;
}
