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>
const int maxn = 1e5 + 10;
using namespace std;
int n;
int col[maxn];
int siz[maxn];
int son[maxn];
int cnt[maxn];
int SON;
int mx;
long long sum = 0;
long long ans[maxn];
vector<int> mp[maxn];
void dfs(int x, int fa) {
  siz[x] = 1;
  for (int v : mp[x]) {
    if (v == fa) continue;
    dfs(v, x);
    siz[x] += siz[v];
    if (siz[v] > siz[son[x]]) son[x] = v;
  }
}
void add(int x, int fa, int val) {
  cnt[col[x]] += val;
  if (cnt[col[x]] > mx) {
    mx = cnt[col[x]];
    sum = col[x];
  } else if (cnt[col[x]] == mx)
    sum += col[x];
  for (int v : mp[x]) {
    if (v == fa || v == SON) continue;
    add(v, x, val);
  }
}
void dfss(int x, int fa, int opt) {
  for (int v : mp[x]) {
    if (v == fa) continue;
    if (v != son[x]) dfss(v, x, 0);
  }
  if (son[x]) {
    dfss(son[x], x, 1);
    SON = son[x];
  }
  add(x, fa, 1);
  SON = 0;
  ans[x] = sum;
  if (!opt) {
    add(x, fa, -1);
    sum = 0;
    mx = 0;
  }
}
int main() {
  ios::sync_with_stdio(false);
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> col[i];
  }
  int u, v;
  for (int i = 1; i < n; i++) {
    cin >> u >> v;
    mp[u].push_back(v);
    mp[v].push_back(u);
  }
  dfs(1, 0);
  dfss(1, 0, 0);
  for (int i = 1; i <= n; i++) cout << ans[i] << " ";
  return 0;
}
