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 mn = 1e5 + 10;
int c[mn], s[mn];
vector<int> g[mn];
map<int, int> *m[mn];
int num[mn];
long long ans[mn];
void dfs(int x, int p) {
  s[x] = 1;
  int bes = 0;
  for (int y : g[x]) {
    if (y == p) continue;
    dfs(y, x);
    s[x] += s[y];
    if (s[y] > s[bes]) bes = y;
  }
  num[x] = num[bes];
  m[x] = m[bes];
  ans[x] = ans[bes];
  if (s[x] == 1) m[x] = new map<int, int>();
  for (int y : g[x]) {
    if (y == p || y == bes) continue;
    for (pair<int, int> th : (*m[y])) {
      int ne = ((*m[x])[th.first] += th.second);
      if (ne > num[x])
        ans[x] = th.first, num[x] = ne;
      else if (ne == num[x])
        ans[x] += th.first;
    }
  }
  int ne = ((*m[x])[c[x]] += 1);
  if (ne > num[x])
    ans[x] = c[x], num[x] = ne;
  else if (ne == num[x])
    ans[x] += c[x];
}
int main() {
  cin.tie(0);
  cin.sync_with_stdio(0);
  int n, i, a, b;
  cin >> n;
  for (i = 1; i <= n; i++) cin >> c[i];
  for (i = 0; i < n - 1; i++)
    cin >> a >> b, g[a].push_back(b), g[b].push_back(a);
  dfs(1, 0);
  for (i = 1; i <= n; i++) printf("%lld ", ans[i]);
}
