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 + 10;
bool mark[MAXN];
vector<long long int> v[MAXN];
map<long long int, long long int> mm[MAXN];
map<long long int, long long int> m;
long long int par[MAXN];
long long int sum[MAXN];
long long int mxx[MAXN];
long long int a[MAXN];
void DFS(long long int x) {
  mark[x] = true;
  if (v[x].size() == 1 && x != 1) {
    m[x] = x;
    mm[x][a[x]]++;
    sum[x] = a[x];
    mxx[x] = 1;
  } else {
    long long int mx = 0;
    long long int pnt = 0;
    for (auto i : v[x]) {
      if (mark[i]) continue;
      par[i] = x;
      DFS(i);
      if (mm[m[i]].size() > mx) {
        mx = mm[m[i]].size();
        pnt = i;
      }
    }
    m[x] = m[pnt];
    sum[x] = sum[pnt];
    mxx[x] = mxx[pnt];
    mm[m[x]][a[x]]++;
    if (mm[m[x]][a[x]] > mxx[x]) {
      mxx[x] = mm[m[x]][a[x]];
      sum[x] = a[x];
    } else if (mm[m[x]][a[x]] == mxx[x]) {
      sum[x] += a[x];
    }
    for (auto i : v[x]) {
      if (par[x] != i && i != pnt) {
        long long int mx = 0;
        for (auto j : mm[m[i]]) {
          mm[m[x]][j.first] += j.second;
          if (mm[m[x]][j.first] > mxx[x]) {
            mxx[x] = mm[m[x]][j.first];
            sum[x] = j.first;
          } else if (mm[m[x]][j.first] == mxx[x]) {
            sum[x] += j.first;
          }
        }
      }
    }
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin.tie(0);
  cout.tie(0);
  long long int n;
  cin >> n;
  for (long long int i = 1; i <= n; i++) cin >> a[i];
  for (long long int i = 1; i < n; i++) {
    long long int x, y;
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  DFS(1);
  for (long long int i = 1; i <= n; i++) cout << sum[i] << " ";
}
