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, c[N];
int head[N], numE = 0;
long long ans[N];
struct E {
  int next, v;
} e[N << 1];
map<int, long long> sum[N];
map<int, int> cnt[N];
void add(int u, int v) {
  e[++numE] = (E){head[u], v};
  head[u] = numE;
}
void dfs(int u, int last) {
  cnt[u][c[u]]++;
  sum[u][1] = c[u];
  for (int i = head[u]; i; i = e[i].next) {
    int v = e[i].v;
    if (v == last) continue;
    dfs(v, u);
    if (cnt[v].size() > cnt[u].size()) {
      swap(cnt[v], cnt[u]);
      swap(sum[v], sum[u]);
    }
    for (map<int, int>::iterator it = cnt[v].begin(); it != cnt[v].end();
         it++) {
      sum[u][cnt[u][it->first]] -= it->first;
      cnt[u][it->first] += it->second;
      sum[u][cnt[u][it->first]] += it->first;
    }
    cnt[v].clear();
    sum[v].clear();
  }
  map<int, long long>::iterator it = sum[u].end();
  it--;
  ans[u] = it->second;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", c + i);
  for (int i = 1, u, v; i < n; i++) {
    scanf("%d%d", &u, &v);
    add(u, v);
    add(v, u);
  }
  dfs(1, 0);
  for (int i = 1; i <= n; i++) printf("%lld ", ans[i]);
  return 0;
}
