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 + 10;
map<int, int> mp[maxn];
int long long n, m, sum[maxn], t[maxn], c[maxn], val[maxn];
vector<int> veci[maxn];
void dfs(int a, int b) {
  for (auto x : veci[a])
    if (x != b) {
      dfs(x, a);
      if (val[a] < val[x]) {
        swap(mp[a], mp[x]);
        swap(t[a], t[x]);
        sum[a] = sum[x];
      }
      val[a] += val[x];
      for (auto y : mp[x]) {
        mp[a][y.first] += y.second;
        if (mp[a][y.first] > t[a]) {
          t[a] = mp[a][y.first];
          sum[a] = 0;
        }
        if (mp[a][y.first] == t[a]) sum[a] += y.first;
      }
      mp[x].clear();
    }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 0; i < n; i++) cin >> c[i];
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b;
    a--;
    b--;
    veci[a].push_back(b);
    veci[b].push_back(a);
  }
  for (int i = 0; i < n; i++) {
    t[i] = 1;
    sum[i] = c[i];
    mp[i][c[i]] = 1;
    val[i] = 1;
  }
  cout << endl;
  dfs(0, 0);
  for (int i = 0; i < n; i++) cout << sum[i] << " ";
}
