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 = 100010;
int n, c[N], pc[N];
long long res[N];
vector<int> E[N];
map<int, int> S[N];
map<int, long long> cnt[N];
void dfs(int cur, int pre) {
  S[cur].insert(make_pair(c[cur], 1));
  cnt[cur].insert(make_pair(1, c[cur]));
  for (int j = (0); j <= (E[cur].size() - 1); j++) {
    int to = E[cur][j];
    if (to == pre) continue;
    dfs(to, cur);
    if (S[pc[to]].size() > S[pc[cur]].size()) swap(pc[to], pc[cur]);
    for (__typeof(S[pc[to]].begin()) it = S[pc[to]].begin();
         it != S[pc[to]].end(); it++) {
      cnt[pc[cur]][S[pc[cur]][it->first]] -= it->first;
      S[pc[cur]][it->first] += it->second;
      cnt[pc[cur]][S[pc[cur]][it->first]] += it->first;
    }
  }
  res[cur] = cnt[pc[cur]].rbegin()->second;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= (n); i++) scanf("%d", &c[i]);
  for (int i = 1; i <= (n - 1); i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    E[u].push_back(v);
    E[v].push_back(u);
  }
  for (int i = 1; i <= (n); i++) pc[i] = i;
  dfs(1, 0);
  for (int i = 1; i <= (n); i++) printf("%I64d ", res[i]);
  printf("\n");
  return 0;
}
