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 long long int maxn = (1e5) + 10, mod = 1e9 + 7;
long long int n, c[maxn], pnt[maxn], res[maxn], num[maxn], ans[maxn], sz[maxn];
vector<long long int> v[maxn], g;
map<long long int, long long int> cnt[maxn];
void DFS2(long long int x, long long int p, long long int r, long long int pn) {
  cnt[pn][c[x]]++;
  if (cnt[pn][c[x]] == res[pn])
    num[pn] += c[x];
  else if (cnt[pn][c[x]] > res[pn]) {
    num[pn] = c[x];
    res[pn] = cnt[pn][c[x]];
  }
  for (auto u : v[x])
    if (u != r && u != p) DFS2(u, x, r, pn);
}
void DFS(long long int x, long long int p) {
  sz[x] = 0;
  pnt[x] = x;
  for (auto u : v[x])
    if (u != p) DFS(u, x);
  for (auto u : v[x])
    if (u != p)
      if (sz[u] > sz[pnt[x]]) pnt[x] = u;
  long long int r = pnt[x];
  pnt[x] = pnt[pnt[x]];
  sz[x] = 1;
  DFS2(x, p, r, pnt[x]);
  for (long long int u : v[x]) {
    if (u == p) continue;
    sz[x] += sz[u];
  }
  ans[x] = num[pnt[x]];
}
int32_t main() {
  ios::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  cin >> n;
  for (long long int i = 1; i <= n; i++) cin >> c[i];
  for (long long int i = 0; i < n - 1; i++) {
    long long int x, y;
    cin >> x >> y;
    v[x].push_back(y);
    v[y].push_back(x);
  }
  DFS(1, -1);
  for (long long int i = 1; i <= n; i++) cout << ans[i] << ' ';
  return 0;
}
