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 N = 100005;
int n, color[N];
long long ans[N];
vector<int> tree[N];
map<int, long long> freq[N], sum[N];
void merge(int u, int v) {
  if (freq[u].size() < freq[v].size())
    swap(freq[u], freq[v]), swap(sum[u], sum[v]);
  for (auto it = freq[v].begin(); it != freq[v].end(); it++) {
    int c = it->first, f = it->second;
    sum[u][freq[u][c]] -= c;
    freq[u][c] += f;
    sum[u][freq[u][c]] += c;
  }
}
void dfs(int u, int parent) {
  freq[u][color[u]] = 1;
  sum[u][1] = color[u];
  for (int i = 0; i < ((int)tree[u].size()); i++) {
    int v = tree[u][i];
    if (parent == v) continue;
    dfs(v, u);
    merge(u, v);
  }
  ans[u] = sum[u].rbegin()->second;
}
int main() {
  scanf("%d", &n);
  for (int i = 0; i < n; i++) scanf("%d", color + i);
  for (int i = 0; i < n - 1; i++) {
    int x, y;
    scanf("%d %d", &x, &y);
    x--, y--;
    tree[x].push_back(y);
    tree[y].push_back(x);
  }
  dfs(0, -1);
  for (int i = 0; i < n; i++)
    printf("%lld%s", ans[i], (i == n - 1) ? "\n" : " ");
  return 0;
}
