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 maxn = 1e5 + 20;
long long n, sz[maxn], cnt[maxn], color[maxn], ans[maxn], sum, maxans;
vector<int> vec[maxn], adj[maxn];
void szbuild(int i, int par) {
  sz[i] = 1;
  for (int x : adj[i])
    if (x != par) szbuild(x, i), sz[i] += sz[x];
}
void dfs(int v, int p, bool keep) {
  int Maxc = -1, bigchild = 0;
  for (auto u : adj[v])
    if (u != p && Maxc < sz[u]) Maxc = sz[u], bigchild = u;
  for (auto u : adj[v])
    if (u != p && u != bigchild) dfs(u, v, 0);
  if (bigchild != 0) dfs(bigchild, v, 1);
  swap(vec[v], vec[bigchild]);
  vec[v].push_back(v);
  cnt[color[v]]++;
  if (cnt[color[v]] == maxans) sum += color[v];
  if (cnt[color[v]] > maxans) maxans = cnt[color[v]], sum = color[v];
  for (auto u : adj[v])
    if (u != p && u != bigchild) {
      for (auto x : vec[u]) {
        cnt[color[x]]++;
        vec[v].push_back(x);
        if (cnt[color[x]] == maxans) sum += color[x];
        if (cnt[color[x]] > maxans) maxans = cnt[color[x]], sum = color[x];
      }
    }
  ans[v] = sum;
  if (keep == 0) {
    for (auto u : vec[v]) cnt[color[u]]--;
    sum = maxans = 0;
  }
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> color[i];
  for (int i = 1, u, v; i < n; i++) {
    cin >> u >> v;
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  szbuild(1, 0);
  dfs(1, 0, 1);
  for (int i = 1; i <= n; i++) cout << ans[i] << ' ';
  cout << endl;
  return 0;
}
