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;
struct node {
  long long l, r, tot, maxx;
} t[100000 * 20];
vector<long long> G[100000 + 5];
long long n, cnt;
long long c[100000 + 5], ans[100000 + 5], root[100000 + 5];
void update(long long now) {
  long long l = t[now].l, r = t[now].r;
  if (t[l].maxx > t[r].maxx)
    t[now].maxx = t[l].maxx, t[now].tot = t[l].tot;
  else if (t[r].maxx > t[l].maxx)
    t[now].maxx = t[r].maxx, t[now].tot = t[r].tot;
  else
    t[now].maxx = t[l].maxx, t[now].tot = t[l].tot + t[r].tot;
}
void modify(long long &now, long long l, long long r, long long pos) {
  if (!now) now = ++cnt;
  if (l == r) {
    t[now].maxx++;
    t[now].tot = l;
    return;
  }
  long long mid = (l + r) >> 1;
  if (pos <= mid)
    modify(t[now].l, l, mid, pos);
  else
    modify(t[now].r, mid + 1, r, pos);
  update(now);
}
long long merge(long long x, long long y, long long l, long long r) {
  if (!x || !y) return x + y;
  if (l == r) {
    t[x].tot = l;
    t[x].maxx += t[y].maxx;
    return x;
  }
  long long mid = (l + r) >> 1;
  t[x].l = merge(t[x].l, t[y].l, l, mid);
  t[x].r = merge(t[x].r, t[y].r, mid + 1, r);
  update(x);
  return x;
}
void dfs(long long u, long long fa) {
  for (long long i = 0; i < G[u].size(); i++) {
    long long v = G[u][i];
    if (v == fa) continue;
    dfs(v, u);
    merge(root[u], root[v], 1, 100000);
  }
  modify(root[u], 1, 100000, c[u]);
  ans[u] = t[root[u]].tot;
}
signed main() {
  scanf("%lld", &n);
  for (long long i = 1; i <= n; i++) {
    scanf("%lld", &c[i]);
    root[i] = i;
    cnt++;
  }
  for (long long i = 1, u, v; i < n; i++) {
    scanf("%lld %lld", &u, &v);
    G[u].push_back(v);
    G[v].push_back(u);
  }
  dfs(1, 0);
  for (long long i = 1; i <= n; i++) printf("%lld ", ans[i]);
  return 0;
}
