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;
template <typename T, typename L>
bool smax(T &x, L y) {
  return x < y ? (x = y, true) : false;
}
template <typename T, typename L>
bool smin(T &x, L y) {
  return y < x ? (x = y, true) : false;
}
void pre_dfs(int, int);
void sack(int, int, bool);
void update(int, int, int, int, int);
const int MAXN = 1e5 + 15;
int cnt[MAXN], col[MAXN];
int st[MAXN], ft[MAXN], add[MAXN];
vector<int> adj[MAXN];
int n;
long long ans[MAXN];
pair<int, long long> tree[MAXN << 2];
int main() {
  ios_base::sync_with_stdio(false), cin.tie(NULL), cout.tie(NULL);
  cerr << "HELLO WORLD :)\n";
  cin >> n;
  for (int i = 0; i < n; i++) cin >> col[i];
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    u--, v--;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  pre_dfs(0, -1);
  sack(0, -1, 0);
  for (int i = 0; i < n; i++) cout << ans[i] << ' ';
  return cout << '\n', false;
}
void pre_dfs(int v, int p) {
  static int t = 0;
  st[v] = t++;
  add[st[v]] = v;
  for (auto u : adj[v])
    if (u != p) pre_dfs(u, v);
  ft[v] = t;
}
void sack(int v, int p, bool keep) {
  int id = -1, mx = -1061109567;
  for (auto u : adj[v])
    if (u != p && smax(mx, ft[u] - st[u])) id = u;
  for (auto u : adj[v])
    if (u != p && u != id) sack(u, v, 0);
  if (id != -1) sack(id, v, 1);
  update(1, 1, n + 1, col[v], 1);
  for (auto u : adj[v])
    if (u != p && u != id)
      for (int i = st[u]; i < ft[u]; i++) update(1, 1, n + 1, col[add[i]], 1);
  ans[v] = tree[1].second;
  if (!keep)
    for (int i = st[v]; i < ft[v]; i++) update(1, 1, n + 1, col[add[i]], -1);
}
void update(int id, int b, int e, int p, int val) {
  if (b == e - 1) {
    tree[id].first += val;
    smax(tree[id].second, b);
    return;
  }
  int mid = (b + e) >> 1;
  if (p < mid)
    update(id << 1, b, mid, p, val);
  else
    update(id << 1 | 1, mid, e, p, val);
  tree[id].first = max(tree[id << 1].first, tree[id << 1 | 1].first);
  tree[id].second = (tree[id << 1].first == tree[id << 1 | 1].first
                         ? tree[id << 1].second + tree[id << 1 | 1].second
                         : (tree[id << 1].first > tree[id << 1 | 1].first
                                ? tree[id << 1].second
                                : tree[id << 1 | 1].second));
}
