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;
int c[maxn + 5];
map<int, int> cnt[maxn + 5];
map<int, long long> sum[maxn + 5];
long long ans[maxn + 5];
int n;
vector<int> g[maxn + 5];
void dfs(int k, int fa) {
  cnt[k][c[k]] = 1;
  sum[k][1] = c[k];
  for (auto u : g[k]) {
    if (u == fa) continue;
    dfs(u, k);
    if (cnt[k].size() < cnt[u].size()) cnt[k].swap(cnt[u]), sum[k].swap(sum[u]);
    for (auto x : cnt[u]) {
      sum[k][cnt[k][x.first]] -= x.first;
      cnt[k][x.first] += x.second;
      sum[k][cnt[k][x.first]] += x.first;
    }
    map<int, int>().swap(cnt[u]);
    map<int, long long>().swap(sum[u]);
  }
  ans[k] = sum[k].rbegin()->second;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &c[i]);
  for (int i = 1; i < n; ++i) {
    int x, y;
    scanf("%d%d", &x, &y);
    g[x].push_back(y), g[y].push_back(x);
  }
  dfs(1, 0);
  for (int i = 1; i <= n; ++i) printf("%lld ", ans[i]);
  printf("\n");
  return 0;
}
