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>
long long read() {
  static char c;
  static long long x;
  int flag = 1;
  while (c = getchar(), !isdigit(c))
    if (c == '-') flag = 0;
  x = c - '0';
  while (c = getchar(), isdigit(c)) x = x * 10 + c - '0';
  return flag ? x : -x;
}
struct edge {
  int pre, to;
} gs[200200];
int head[100100], gc;
void add(int x, int y) {
  gs[++gc] = (edge){head[x], y};
  head[x] = gc;
}
int size[100100], son[100100], fa[100100], ids[100100], left[100100],
    pos[100100];
void count(int x) {
  static int index = 0;
  ids[x] = ++index;
  pos[ids[x]] = x;
  size[x] = 1;
  for (int i = head[x]; i; i = gs[i].pre) {
    if (gs[i].to == fa[x]) continue;
    fa[gs[i].to] = x;
    count(gs[i].to);
    size[x] += size[gs[i].to];
    if (size[gs[i].to] > size[son[x]]) son[x] = gs[i].to;
  }
  left[x] = index;
}
long long ans[100100];
int n, ns[100100];
int maxcnt;
std::unordered_map<int, int> cnt;
void update(int x, int v) {
  ++cnt[v];
  if (cnt[v] > maxcnt) return ans[x] = v, maxcnt = cnt[v], void();
  if (cnt[v] == maxcnt) return ans[x] += v, void();
}
void dfs(int x, bool save = 1) {
  if (!x) return;
  for (int i = head[x]; i; i = gs[i].pre) {
    if (gs[i].to == fa[x] || gs[i].to == son[x]) continue;
    dfs(gs[i].to, 0);
  }
  dfs(son[x], 1);
  ans[x] = ans[son[x]];
  update(x, ns[x]);
  for (int i = head[x]; i; i = gs[i].pre) {
    if (gs[i].to == fa[x] || gs[i].to == son[x]) continue;
    int to = gs[i].to;
    for (int j = ids[to]; j <= left[to]; ++j) {
      int y = pos[j];
      update(x, ns[y]);
    }
  }
  if (!save) cnt.clear(), maxcnt = 0;
}
signed main() {
  n = read();
  for (int i = 1; i <= n; ++i) ns[i] = read();
  for (int i = 1; i < n; ++i) {
    int x = read(), y = read();
    add(x, y);
    add(y, x);
  }
  count(1);
  dfs(1);
  for (int i = 1; i <= n; ++i) printf("%lld ", ans[i]);
}
