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;
int val[100010], fa[100010], id[100010], Max[100010];
long long ans[100010], sum[100010];
vector<int> edge[100010];
map<int, int> cnt[100010];
void merge(int &x, int &y) {
  if (cnt[x].size() < cnt[y].size()) swap(x, y);
  int a, b, c;
  for (map<int, int>::iterator iter = cnt[y].begin(); iter != cnt[y].end();
       ++iter) {
    a = iter->first;
    b = iter->second;
    if (cnt[x].find(a) == cnt[x].end()) cnt[x][a] = 0;
    c = (cnt[x][a] += b);
    if (c > Max[x]) {
      Max[x] = c;
      sum[x] = a;
    } else if (c == Max[x])
      sum[x] += a;
  }
}
void dfs(int i) {
  id[i] = i;
  cnt[i][val[i]] = 1;
  sum[i] = val[i];
  Max[i] = 1;
  for (vector<int>::iterator j = edge[i].begin(); j != edge[i].end(); ++j)
    if (*j != fa[i]) {
      fa[*j] = i;
      dfs(*j);
      merge(id[i], id[*j]);
    }
  ans[i] = sum[id[i]];
}
int main() {
  int n, x, y;
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", val + i);
  for (int i = 1; i < n; ++i) {
    scanf("%d%d", &x, &y);
    edge[x].push_back(y);
    edge[y].push_back(x);
  }
  dfs(1);
  for (int i = 1; i <= n; ++i) printf("%I64d%c", ans[i], i == n ? '\n' : ' ');
  return 0;
}
