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 MAX = 100002;
map<int, int> cnt[MAX];
vector<int> adj[MAX];
long long ans[MAX];
int color[MAX];
int max_el[MAX];
void dfs(int node, int par) {
  cnt[node][color[node]] = true;
  max_el[node] = 1;
  ans[node] = color[node];
  for (int i = 0; i < adj[node].size(); i++) {
    int child = adj[node][i];
    if (child != par) {
      dfs(child, node);
      if (cnt[node].size() < cnt[child].size()) {
        swap(cnt[node], cnt[child]);
        max_el[node] = max_el[child];
        ans[node] = ans[child];
      }
      for (pair<int, int> p : cnt[child]) {
        cnt[node][p.first] += p.second;
        if (cnt[node][p.first] == max_el[node])
          ans[node] += p.first;
        else if (cnt[node][p.first] > max_el[node]) {
          max_el[node] = cnt[node][p.first];
          ans[node] = p.first;
        }
      }
    }
  }
}
int main() {
  int n, u0, v0;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> color[i];
  }
  for (int i = 0; i < n - 1; i++) {
    cin >> u0 >> v0;
    adj[u0].push_back(v0);
    adj[v0].push_back(u0);
  }
  dfs(1, 0);
  for (int i = 1; i <= n; i++) {
    cout << ans[i] << " ";
  }
}
