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;
vector<long long> edge[200006];
long long l[200005], r[200006], a[200006], num[200005], tot = 0;
long long son[200006], sz[200006], ans[200005], pos[200005], mx = 0, sum = 0;
void upd(long long x) {
  if (x == mx) {
    num[x]++;
    sum = x;
  } else {
    num[x]++;
    if (num[mx] == num[x]) sum += x;
    if (num[x] > num[mx]) sum = x, mx = x;
  }
}
void dfs(long long u, long long f) {
  l[u] = ++tot;
  sz[u] = 1;
  for (auto v : edge[u]) {
    if (v == f) continue;
    dfs(v, u);
    sz[u] += sz[v];
    if (sz[son[u]] < sz[v]) son[u] = v;
  }
  r[u] = tot;
}
void cal(long long u, long long fa, long long t) {
  for (auto v : edge[u]) {
    if (v == fa || v == son[u]) continue;
    cal(v, u, 0);
  }
  if (son[u]) cal(son[u], u, 1);
  upd(a[u]);
  for (auto v : edge[u]) {
    if (v == fa || v == son[u]) continue;
    for (int j = l[v]; j <= r[v]; j++) {
      upd(a[pos[j]]);
    }
  }
  ans[u] = sum;
  if (!t) {
    mx = 0;
    sum = 0;
    for (int i = l[u]; i <= r[u]; i++) {
      num[a[pos[i]]]--;
    }
  }
}
int main() {
  int n;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> a[i];
  }
  for (int i = 2; i <= n; i++) {
    int u, v;
    cin >> u >> v;
    edge[u].push_back(v);
    edge[v].push_back(u);
  }
  dfs(1, 0);
  for (int i = 1; i <= n; i++) {
    pos[l[i]] = i;
  }
  cal(1, 0, 1);
  for (int i = 1; i <= n; i++) {
    cout << ans[i] << " ";
  }
  cout << endl;
}
