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 = 1000 * 200;
int color[MAXN];
long long sum[MAXN], sz[MAXN];
vector<int> g[MAXN];
int was[MAXN];
typedef map<long long, long long>* pmap;
pmap m[MAXN];
pmap merge(pmap a, pmap b, long long& sum, long long& sz) {
  if (a->size() > b->size()) {
    return merge(b, a, sum, sz);
  }
  for (auto it : *a) {
    (*b)[it.first] += it.second;
    long long buf_sz = (*b)[it.first];
    if (sz < buf_sz) {
      sz = buf_sz;
      sum = it.first;
    } else {
      if (sz == buf_sz) {
        sum += it.first;
      }
    }
  }
  return b;
}
void dfs(int v) {
  m[v] = new map<long long, long long>();
  was[v] = 1;
  (*m[v])[color[v]] = 1;
  sum[v] = color[v];
  sz[v] = 1;
  for (int u : g[v]) {
    if (!was[u]) {
      dfs(u);
      if (m[v]->size() < m[u]->size()) {
        sum[v] = sum[u];
        sz[v] = sz[u];
      }
      m[v] = merge(m[u], m[v], sum[v], sz[v]);
    }
  }
}
int main() {
  ios_base::sync_with_stdio(false);
  cin.tie(0);
  int n;
  cin >> n;
  for (int i = 0; i < n; i++) cin >> color[i];
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    cin >> u >> v;
    u--;
    v--;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(0);
  for (int i = 0; i < n; i++) {
    cout << sum[i] << "\n";
  }
  return 0;
}
