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;
map<int, int> *cnt[100010];
vector<int> g[100010];
int col[100010];
long long res[100010], ma[100010];
int n;
void push(int v, int c) {
  if (ma[v] < (*cnt[v])[c]) {
    res[v] = c, ma[v] = (*cnt[v])[c];
  } else if (ma[v] == (*cnt[v])[c]) {
    res[v] += c;
  }
}
int dfs(int v, int p) {
  long long sz = 0;
  ma[v] = 0, res[v] = 0;
  long long msz = -1, mv = -1;
  for (auto to : g[v]) {
    if (to != p) {
      long long nsz = dfs(to, v);
      sz += nsz;
      if (nsz > msz) msz = nsz, mv = to;
    }
  }
  if (mv != -1)
    cnt[v] = cnt[mv], res[v] = res[mv], ma[v] = ma[mv];
  else
    cnt[v] = new map<int, int>();
  ++(*cnt[v])[col[v]];
  push(v, col[v]);
  for (auto to : g[v]) {
    if (to != p && to != mv) {
      for (auto x : *cnt[to]) {
        if (x.second == 0) continue;
        (*cnt[v])[x.first] += x.second;
        push(v, x.first);
      }
    }
  }
  return sz + 1;
}
int main() {
  cin >> n;
  for (int i = 1; i < n + 1; ++i) cin >> col[i];
  for (int i = 0; i < n - 1; ++i) {
    int a, b;
    cin >> a >> b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  dfs(1, -1);
  for (int i = 1; i < n + 1; ++i) cout << res[i] << " ";
}
