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 long long inf = 1e5 + 9;
long long n, c[inf], sz[inf], MaxFrequency[inf], ans[inf];
vector<long long> adj[inf];
map<long long, long long> *Cnt[inf], *SumCnt[inf];
void dfs(long long u, long long p) {
  auto add = [&](long long color, long long val) {
    (*SumCnt[u])[(*Cnt[u])[color]] -= color;
    (*Cnt[u])[color] += val;
    (*SumCnt[u])[(*Cnt[u])[color]] += color;
    MaxFrequency[u] = max(MaxFrequency[u], (*Cnt[u])[color]);
  };
  sz[u] = 1;
  long long mx = 0, heavy = -1;
  for (auto v : adj[u]) {
    if (v == p) continue;
    dfs(v, u);
    sz[u] += sz[v];
    if (sz[v] > mx) mx = sz[v], heavy = v;
  }
  if (heavy == -1)
    Cnt[u] = new map<long long, long long>(),
    SumCnt[u] = new map<long long, long long>(), MaxFrequency[u] = 0;
  else
    Cnt[u] = Cnt[heavy], SumCnt[u] = SumCnt[heavy],
    MaxFrequency[u] = MaxFrequency[heavy];
  add(c[u], 1);
  for (auto v : adj[u]) {
    if (v == p || v == heavy) continue;
    for (auto o : *Cnt[v]) add(o.first, o.second);
  }
  ans[u] = (*SumCnt[u])[MaxFrequency[u]];
}
int main() {
  cin >> n;
  for (long long i = 1; i <= n; i++) cin >> c[i];
  for (long long i = 1; i < n; i++) {
    long long a, b;
    cin >> a >> b;
    adj[a].push_back(b);
    adj[b].push_back(a);
  }
  dfs(1, 0);
  for (long long i = 1; i <= n; i++) cout << ans[i] << " ";
  cout << endl;
}
