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;
bool debug = 0;
int n, m, k;
int dx[4] = {0, 1, 0, -1}, dy[4] = {1, 0, -1, 0};
long long ln, lk, lm;
vector<int> mp[100005];
int c[100005];
long long ans[100005];
unordered_map<int, int> tMap[100005];
long long tmpAns[100005];
int cnt[100005], id[100005];
void mg(int &x, int y) {
  if (tMap[x].size() < tMap[y].size()) swap(x, y);
  for (auto p : tMap[y]) {
    tMap[x][p.first] += p.second;
    if (tMap[x][p.first] > cnt[x]) {
      tmpAns[x] = p.first;
      cnt[x] = tMap[x][p.first];
    } else if (tMap[x][p.first] == cnt[x]) {
      tmpAns[x] += p.first;
    }
  }
}
void dfs(int x, int pa) {
  id[x] = x;
  tMap[x][c[x]] = 1;
  tmpAns[id[x]] = c[x];
  cnt[id[x]] = 1;
  for (int v : mp[x])
    if (v != pa) {
      dfs(v, x);
      mg(id[x], id[v]);
    }
  ans[x] = tmpAns[id[x]];
}
void fmain() {
  scanf("%d", &n);
  for (int(i) = 1; (i) <= (int)(n); (i)++) scanf("%d", c + i);
  for (int u, v, i = 0; i < n - 1; i++) {
    scanf("%d%d", &u, &v);
    mp[u].push_back(v);
    mp[v].push_back(u);
  }
  dfs(1, 0);
  for (int(i) = 1; (i) <= (int)(n); (i)++) printf("%lld ", ans[i]);
}
int main() {
  fmain();
  return 0;
}
