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 maxn = 1e5 + 7;
struct Tree {
  int lc, rc;
  long long sum, ans;
} t[5000050];
int rt[maxn], col[maxn], n;
long long ans[maxn];
int head[maxn], cnt = 1, tot;
struct Node {
  int v, nxt;
} G[maxn << 1];
void ins(int u, int v) {
  G[cnt] = (Node){v, head[u]};
  head[u] = cnt++;
}
struct Segment_Tree {
  void update(int o) {
    if (t[t[o].lc].sum > t[t[o].rc].sum)
      t[o].sum = t[t[o].lc].sum, t[o].ans = t[t[o].lc].ans;
    else if (t[t[o].lc].sum < t[t[o].rc].sum)
      t[o].sum = t[t[o].rc].sum, t[o].ans = t[t[o].rc].ans;
    else
      t[o].sum = t[t[o].lc].sum, t[o].ans = t[t[o].lc].ans + t[t[o].rc].ans;
  }
  void Modify(int &o, int l, int r, int ql) {
    if (!o) o = ++tot;
    if (l == r) {
      t[o].ans = l;
      ++t[o].sum;
      return;
    }
    if (ql <= ((l + r) >> 1))
      Modify(t[o].lc, l, ((l + r) >> 1), ql);
    else
      Modify(t[o].rc, ((l + r) >> 1) + 1, r, ql);
    update(o);
  }
};
Segment_Tree T;
int merge(int a, int b, int l, int r) {
  if (!a) return b;
  if (!b) return a;
  if (l == r) {
    t[a].ans = l;
    t[a].sum += t[b].sum;
    return a;
  }
  t[a].lc = merge(t[a].lc, t[b].lc, l, ((l + r) >> 1));
  t[a].rc = merge(t[a].rc, t[b].rc, ((l + r) >> 1) + 1, r);
  T.update(a);
  return a;
}
void dfs(int x, int fa) {
  for (int i = head[x]; i; i = G[i].nxt) {
    int v = G[i].v;
    if (v == fa) continue;
    dfs(v, x);
    merge(rt[x], rt[v], 1, 100000);
  }
  T.Modify(rt[x], 1, 100000, col[x]);
  ans[x] = t[rt[x]].ans;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) {
    scanf("%d", &col[i]);
    rt[i] = i;
    ++tot;
  }
  for (int i = 1; i <= n - 1; ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    ins(u, v);
    ins(v, u);
  }
  dfs(1, 0);
  for (int i = 1; i <= n; ++i) printf("%lld ", ans[i]);
  return 0;
}
