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;
int n;
int c[MAXN];
vector<int> E[MAXN];
map<int, int> *lst[MAXN];
int cnt[MAXN];
long long sum[MAXN];
void insert(int u, pair<int, int> p) {
  (*lst[u])[p.first] += p.second;
  if ((*lst[u])[p.first] > cnt[u])
    cnt[u] = (*lst[u])[p.first], sum[u] = p.first;
  else if ((*lst[u])[p.first] == cnt[u])
    sum[u] += p.first;
}
void DFS(int u, int pre) {
  int w = 0;
  for (int v : E[u])
    if (v != pre) {
      DFS(v, u);
      if (!w || lst[w]->size() < lst[v]->size()) w = v;
    }
  if (w) {
    lst[u] = lst[w];
    cnt[u] = cnt[w];
    sum[u] = sum[w];
    for (auto v : E[u])
      if (v != pre && v != w) {
        for (auto p : *lst[v]) insert(u, p);
        delete lst[v];
      }
  } else
    lst[u] = new map<int, int>;
  insert(u, make_pair(c[u], 1));
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &c[i]);
  for (int u, v, i = 1; i < n; ++i) {
    scanf("%d%d", &u, &v);
    E[u].push_back(v);
    E[v].push_back(u);
  }
  DFS(1, 0);
  for (int i = 1; i <= n; ++i) printf("%I64d%c", sum[i], "\n "[i < n]);
  return 0;
}
