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 Mod = 7 + 1e9;
const long long Mod2 = 1 + 1e7;
const long long INF32 = 5 + 2e9;
const long long INF64 = 5 + 1e18;
const long long MAX = 5 + 2e5;
long long n, C[MAX], cnt[MAX], sum[MAX], MX, sz[MAX];
bool Big[MAX];
vector<long long> G[MAX];
int GetSz(int u, int p) {
  sz[u] = 1;
  for (auto v : G[u])
    if (v != p) sz[u] += GetSz(v, u);
  return sz[u];
}
void add(int u, int p, int x) {
  long long &CNT = cnt[C[u]];
  sum[CNT] -= C[u];
  CNT += x;
  sum[CNT] += C[u];
  if (sum[MX + 1]) MX++;
  if (!sum[MX]) MX--;
  for (auto v : G[u])
    if (v != p && !Big[v]) add(v, u, x);
}
long long ans[MAX];
void DFS(int u, int p, bool keep) {
  int B = 0;
  for (auto v : G[u])
    if (v != p && sz[v] > sz[B]) B = v;
  Big[B] = 1;
  for (auto v : G[u])
    if (v != p && v != B) DFS(v, u, 0);
  if (B) DFS(B, u, 1);
  add(u, p, 1);
  ans[u] = sum[MX];
  Big[B] = 0;
  if (!keep) add(u, p, -1);
}
int main() {
  ios::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  cin >> n;
  for (int i = 1; i <= n; i++) cin >> C[i];
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    G[u].push_back(v);
    G[v].push_back(u);
  }
  GetSz(1, 1);
  DFS(1, 1, 1);
  for (int i = 1; i <= n; i++) cout << ans[i] << ' ';
  cout << endl;
  return 0;
}
