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>
int n, id[100005], maxt[100005], c[100005], eh[100005], ep;
long long sum[100005], ans[100005];
std::map<int, int> ct[100005];
class edge {
 public:
  int v, next;
} e[100005 << 1];
void add_edge(int u, int v) {
  e[ep].v = v;
  e[ep].next = eh[u];
  eh[u] = ep++;
  e[ep].v = u;
  e[ep].next = eh[v];
  eh[v] = ep++;
}
int merge(int x, int y) {
  int t;
  if (ct[x].size() < ct[y].size()) std::swap(x, y);
  for (std::map<int, int>::iterator it = ct[y].begin(); it != ct[y].end();
       ++it) {
    t = (ct[x][it->first] += it->second);
    if (maxt[x] < t) {
      maxt[x] = t;
      sum[x] = it->first;
    } else if (maxt[x] == t)
      sum[x] += (long long)(it->first);
  }
  ct[y].clear();
  return x;
}
void dfs(int x, int p) {
  id[x] = x;
  ct[x][c[x]] = maxt[x] = 1;
  sum[x] = c[x];
  for (int i = eh[x]; i != -1; i = e[i].next) {
    if (e[i].v == p) continue;
    dfs(e[i].v, x);
    id[x] = merge(id[x], id[e[i].v]);
  }
  ans[x] = sum[id[x]];
}
int main() {
  scanf("%d", &n);
  memset(eh, -1, sizeof(eh));
  for (int i = 1; i <= n; ++i) scanf("%d", c + i);
  for (int i = 0, u, v; i < n - 1; ++i) {
    scanf("%d%d", &u, &v);
    add_edge(u, v);
  }
  dfs(1, 0);
  for (int i = 1; i < n; ++i) printf("%I64d ", ans[i]);
  printf("%I64d\n", ans[n]);
  return 0;
}
