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 long long N = 100010;
long long n, m, tot, cnt, head[N], root[N], b[N], ans[N];
struct Edge {
  long long suiv, ver;
} e[N << 1];
struct SegmentTree {
  long long l, r, sum, maxi;
} t[N * 40];
inline void lnk(long long x, long long y) {
  e[++tot].ver = y;
  e[tot].suiv = head[x];
  head[x] = tot;
}
inline void update(long long p) {
  if (t[t[p].l].maxi == t[t[p].r].maxi)
    t[p].sum = t[t[p].l].sum + t[t[p].r].sum;
  else
    t[p].sum = t[t[p].l].maxi <= t[t[p].r].maxi ? t[t[p].r].sum : t[t[p].l].sum;
  t[p].maxi = max(t[t[p].l].maxi, t[t[p].r].maxi);
}
inline void insert(long long &p, long long l, long long r, long long d) {
  if (!p) p = ++cnt;
  if (l == r) {
    t[p].sum += l, t[p].maxi++;
    return;
  }
  long long mid = l + r >> 1;
  if (d <= mid) insert(t[p].l, l, mid, d);
  if (d > mid) insert(t[p].r, mid + 1, r, d);
  update(p);
}
inline long long merge(long long p, long long q, long long l, long long r) {
  if (!p || !q) return p ^ q;
  if (l == r) {
    t[p].sum = t[q].sum, t[p].maxi += t[q].maxi;
    return p;
  }
  long long mid = l + r >> 1;
  t[p].l = merge(t[p].l, t[q].l, l, mid);
  t[p].r = merge(t[p].r, t[q].r, mid + 1, r);
  update(p);
  return p;
}
inline void dfs(long long x, long long fa) {
  for (long long i = head[x]; i; i = e[i].suiv) {
    long long y = e[i].ver;
    if (y == fa) continue;
    dfs(y, x);
    root[x] = merge(root[x], root[y], 1, n);
  }
  ans[x] = t[root[x]].sum;
}
signed main() {
  scanf("%lld", &n);
  for (long long i = 1, pos; i <= n; i++)
    scanf("%lld", &pos), insert(root[i], 1, n, pos);
  for (long long i = 1, x, y; i < n; i++)
    scanf("%lld%lld", &x, &y), lnk(x, y), lnk(y, x);
  dfs(1, 1);
  for (long long i = 1; i <= n; i++) printf("%lld ", ans[i]);
  return 0;
}
