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;
long long n, cols[MAXN], ans[MAXN];
vector<int> adj[MAXN];
unordered_map<long long, long long> c1[MAXN];
map<long long, long long> sm[MAXN];
void merge(int c, int nxt) {
  if (c1[c].size() < c1[nxt].size()) c1[c].swap(c1[nxt]), sm[c].swap(sm[nxt]);
  for (auto it : c1[nxt]) {
    long long oldAmt = c1[c][it.first];
    if (oldAmt) sm[c][oldAmt] -= it.first;
    c1[c][it.first] += it.second;
    long long newAmt = oldAmt + it.second;
    sm[c][newAmt] += it.first;
  }
  c1[nxt].clear();
  sm[nxt].clear();
}
void dfs(int c, int p) {
  c1[c][cols[c]]++;
  sm[c][1] += cols[c];
  for (auto nxt : adj[c])
    if (nxt != p) {
      dfs(nxt, c);
      merge(c, nxt);
    }
  ans[c] = sm[c].rbegin()->second;
}
int main() {
  cin >> n;
  for (int i = 0; i < n; i++) cin >> cols[i];
  for (int i = 0; i < n - 1; i++) {
    int a, b;
    cin >> a >> b, --a, --b;
    adj[a].push_back(b), adj[b].push_back(a);
  }
  dfs(0, -1);
  for (int i = 0; i < n; i++) cout << ans[i] << " ";
}
