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 = 100010;
int n;
map<int, int> cnt[MAXN];
map<int, long long> sumv[MAXN];
vector<int> G[MAXN];
long long ans[MAXN];
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(sumv[u], sumv[v]);
    }
    for (map<int, int>::iterator it = cnt[v].begin(); it != cnt[v].end();
         it++) {
      sumv[u][cnt[u][it->first]] -= it->first;
      cnt[u][it->first] += it->second;
      sumv[u][cnt[u][it->first]] += it->first;
    }
  }
  ans[u] = sumv[u].rbegin()->second;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) {
    int c;
    scanf("%d", &c);
    cnt[i][c] = 1;
    sumv[i][1] = c;
  }
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    G[u].push_back(v);
    G[v].push_back(u);
  }
  dfs(1, 0);
  for (int i = 1; i <= n; i++) printf("%I64d ", ans[i]);
  return 0;
}
