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 = 200010;
long long n, c[MAX], ans[MAX];
map<int, int> col[MAX];
map<int, long long> sum[MAX];
vector<int> e[MAX];
void merg(int u, int v) {
  if (col[u].size() < col[v].size()) {
    swap(col[u], col[v]);
    swap(sum[u], sum[v]);
  }
  for (map<int, int>::iterator it = col[v].begin(); it != col[v].end(); it++) {
    sum[u][col[u][it->first]] -= it->first;
    col[u][it->first] += col[v][it->first];
    sum[u][col[u][it->first]] += it->first;
  }
}
void dfs(int u, int fa) {
  col[u][c[u]] = 1;
  sum[u][1] = c[u];
  for (int i = 0; i < e[u].size(); i++) {
    int v = e[u][i];
    if (v == fa) {
      continue;
    }
    dfs(v, u);
    merg(u, v);
  }
  ans[u] = sum[u].rbegin()->second;
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> c[i];
    e[i].clear();
  }
  for (int i = 1, x, y; i < n; i++) {
    cin >> x >> y;
    e[x].push_back(y);
    e[y].push_back(x);
  }
  dfs(1, -1);
  for (int i = 1; i <= n; i++) {
    cout << ans[i] << ' ';
  }
  return 0;
}
