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 N = (int)1e5 + 3;
int c[N], ct[N], pt[N];
long long ans[N];
map<int, int> mp[N];
vector<int> g[N];
void merge(int a, int b) {
  if (mp[pt[a]].size() > mp[pt[b]].size()) {
    swap(pt[a], pt[b]);
    ans[b] = ans[a];
  }
  for (auto it : mp[pt[a]]) {
    mp[pt[b]][it.first] += it.second;
    if (mp[pt[b]][it.first] == ct[pt[b]]) ans[b] += it.first;
    if (mp[pt[b]][it.first] > ct[pt[b]]) {
      ct[pt[b]] = mp[pt[b]][it.first];
      ans[b] = it.first;
    }
  }
}
void dfs(int v, int p) {
  for (int i = 0; i < g[v].size(); i++) {
    if (g[v][i] != p) {
      dfs(g[v][i], v);
      merge(g[v][i], v);
    }
  }
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &c[i]);
    pt[i] = i;
    mp[i][c[i]] = 1;
    ct[i] = 1;
    ans[i] = c[i];
  }
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1, 0);
  for (int i = 1; i <= n; i++) {
    if (i > 1) printf(" ");
    printf("%lld", ans[i]);
  }
  cout << endl;
  return 0;
}
