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>
const int N = 200005;
using namespace std;
int n;
int c[N];
map<int, int> S[N];
int w[N];
int used[N];
int mx[N];
long long cnt[N];
vector<int> g[N];
long long ans[N];
void run_merge(int a, int b) {
  if (S[w[a]].size() < S[w[b]].size()) {
    swap(w[a], w[b]);
  }
  int id = w[a];
  map<int, int>::iterator it;
  for (it = S[w[b]].begin(); it != S[w[b]].end(); it++) {
    int val = (*it).first;
    int amount = (*it).second;
    S[id][val] += amount;
    if (S[id][val] > mx[id]) {
      mx[id] = S[id][val];
      cnt[id] = 0;
    }
    if (S[id][val] == mx[id]) cnt[id] += val;
  }
}
void dfs(int v) {
  used[v] = 1;
  for (int i = 0; i < g[v].size(); i++) {
    int to = g[v][i];
    if (used[to]) continue;
    dfs(to);
    run_merge(v, to);
  }
  ans[v] = cnt[w[v]];
}
int main() {
  ios_base::sync_with_stdio(0);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> c[i];
    S[i][c[i]] = 1;
    w[i] = i;
    mx[i] = 1;
    cnt[i] = c[i];
  }
  for (int i = 1; i < n; i++) {
    int a, b;
    cin >> a >> b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  dfs(1);
  for (int i = 1; i <= n; i++) {
    if (i > 1) cout << " ";
    cout << ans[i];
  }
  cout << endl;
  return 0;
}
