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;
using ll = long long;
const int MAXN = 100010;
int N;
int C[MAXN];
vector<int> adj[MAXN];
ll ans[MAXN];
struct subtree {
  map<int, int> mp;
  pair<int, ll> best;
  subtree(int c) : mp({{c, 1}}), best(1, ll(c)) {}
};
void combine(subtree* a, subtree* b) {
  if (a->mp.size() < b->mp.size()) {
    swap(*a, *b);
  }
  for (auto p : b->mp) {
    int& val = a->mp[p.first];
    val += p.second;
    if (val > a->best.first) {
      a->best = {val, 0};
    }
    if (val == a->best.first) {
      a->best.second += p.first;
    }
  }
}
subtree* dfs(int cur = 1, int prv = 0) {
  auto res = new subtree(C[cur]);
  for (int nxt : adj[cur]) {
    if (nxt == prv) continue;
    combine(res, dfs(nxt, cur));
  }
  ans[cur] = res->best.second;
  return res;
}
int main() {
  scanf("%d", &N);
  for (int i = 1; i <= N; i++) {
    scanf("%d", &C[i]);
  }
  for (int e = 0; e < N - 1; e++) {
    int a, b;
    scanf("%d %d", &a, &b);
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  dfs();
  for (int i = 1; i <= N; i++) {
    printf("%lld%c", ans[i], " \n"[i == N]);
  }
  return 0;
}
