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;
int main() {
  ios_base ::sync_with_stdio(false);
  cin.tie(nullptr);
  int n;
  long long maxCnt = 0, tot = 0;
  cin >> n;
  vector<vector<int>> adjList(n, vector<int>());
  vector<long long> size(n), cnt(n + 1), ans(n), c(n);
  vector<bool> big(n, false);
  for (long long &x : c) cin >> x;
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    adjList[--u].push_back(--v);
    adjList[v].push_back(u);
  }
  function<int(int, int)> getSize = [&](int u, int p) -> int {
    size[u] = 1;
    for (int v : adjList[u])
      if (v != p) size[u] += getSize(v, u);
    return size[u];
  };
  getSize(0, -1);
  function<void(int, int, int)> add = [&](int u, int p, int x) {
    cnt[c[u]] += x;
    if (cnt[c[u]] > maxCnt)
      maxCnt = cnt[c[u]], tot = c[u];
    else if (cnt[c[u]] == maxCnt)
      tot += c[u];
    for (int v : adjList[u])
      if (v != p && !big[v]) add(v, u, x);
  };
  function<void(int, int, bool)> dfs = [&](int u, int p, bool keep) {
    int max = -1, bigChild = -1;
    for (int v : adjList[u])
      if (v != p && max < size[v]) max = size[bigChild = v];
    for (int v : adjList[u])
      if (v != p && v != bigChild) dfs(v, u, false);
    if (bigChild != -1) dfs(bigChild, u, true), big[bigChild] = true;
    add(u, p, 1);
    ans[u] = tot;
    if (bigChild != -1) big[bigChild] = false;
    if (!keep) add(u, p, -1), tot = 0, maxCnt = 0;
  };
  dfs(0, -1, false);
  for (int i = 0; i < n; i++) {
    if (i > 0) cout << ' ';
    cout << ans[i];
  }
  cout << '\n';
  return 0;
}
