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;
int n, cnt;
int col[100010], ehead[100010];
long long ans[100010];
struct ed {
  int to, nxt;
} e[100010 << 1];
map<int, int> mp[100010];
map<int, long long> sum[100010];
map<int, int>::iterator it1, it2;
inline int read() {
  int x = 0;
  char ch = getchar();
  while (ch < '0' || ch > '9') ch = getchar();
  while (ch >= '0' && ch <= '9') x = x * 10 + ch - '0', ch = getchar();
  return x;
}
void prework() {
  for (int i = 1; i <= n; i++)
    col[i] = read(), mp[i][col[i]] = 1, sum[i][1] = col[i];
  int u, v;
  cnt = 0;
  for (int i = 1; i <= n - 1; i++) {
    u = read();
    v = read();
    e[++cnt].to = v;
    e[cnt].nxt = ehead[u];
    ehead[u] = cnt;
    e[++cnt].to = u;
    e[cnt].nxt = ehead[v];
    ehead[v] = cnt;
  }
}
void dfs(int u, int fa) {
  for (int i = ehead[u]; i; i = e[i].nxt) {
    int v = e[i].to;
    if (v != fa) {
      dfs(v, u);
      if (mp[u].size() < mp[v].size()) swap(mp[u], mp[v]), swap(sum[u], sum[v]);
      for (it1 = mp[v].begin(); it1 != mp[v].end(); ++it1) {
        sum[u][mp[u][(*it1).first]] -= (*it1).first;
        mp[u][(*it1).first] += (*it1).second;
        sum[u][mp[u][(*it1).first]] += (*it1).first;
      }
    }
  }
  ans[u] = (*sum[u].rbegin()).second;
}
void mainwork() { dfs(1, 0); }
void print() {
  for (int i = 1; i <= n; i++) printf("%I64d ", ans[i]);
  printf("\n");
}
int main() {
  while (~scanf("%d", &n)) {
    prework();
    mainwork();
    print();
  }
  return 0;
}
