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;
template <typename T>
inline bool chkmin(T &a, const T &b) {
  return a > b ? a = b, 1 : 0;
}
template <typename T>
inline bool chkmax(T &a, const T &b) {
  return a < b ? a = b, 1 : 0;
}
const int MAX_N = 100005;
int n;
int c[MAX_N];
int x, y;
vector<int> G[MAX_N];
long long ans[MAX_N];
map<int, int> cnt[MAX_N];
map<int, long long> sum[MAX_N];
map<int, int>::iterator it;
void dfs(int u, int fa) {
  for (int i = 0; i < G[u].size(); i++) {
    int v = G[u][i];
    if (v == fa) continue;
    dfs(v, u);
    if (cnt[u].size() < cnt[v].size()) {
      swap(cnt[u], cnt[v]);
      swap(sum[u], sum[v]);
    }
    for (it = cnt[v].begin(); it != cnt[v].end(); ++it) {
      sum[u][cnt[u][it->first]] -= it->first;
      cnt[u][it->first] += it->second;
      sum[u][cnt[u][it->first]] += it->first;
    }
  }
  ans[u] = sum[u].rbegin()->second;
}
int main() {
  scanf("%d", &n);
  for (int i = (1), _end_ = (n + 1); i < _end_; ++i) {
    scanf("%d", &c[i]);
    cnt[i][c[i]] = 1;
    sum[i][1] = c[i];
  }
  for (int i = (0), _end_ = (n - 1); i < _end_; ++i) {
    scanf("%d%d", &x, &y);
    G[x].emplace_back(y);
    G[y].emplace_back(x);
  }
  dfs(1, 0);
  for (int i = (1), _end_ = (n + 1); i < _end_; ++i)
    printf("%lld%c", ans[i], i == n ? '\n' : ' ');
  return 0;
}
