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;
inline long long read() {
  long long x = 0;
  bool f = 0;
  char c = getchar();
  for (; !isdigit(c); c = getchar()) f ^= !(c ^ 45);
  for (; isdigit(c); c = getchar()) x = (x * 10) + (c ^ 48);
  if (f) x = -x;
  return x;
}
struct abc {
  long long v, nt;
} e[200050];
long long n, m, cnt, head[100050], c[100500], s[100004], son[100050], sn,
    p[100050], mx, sum, ans[100050];
void add(long long u, long long v) {
  e[++cnt].v = v;
  e[cnt].nt = head[u];
  head[u] = cnt;
}
void dfs1(long long x, long long f) {
  s[x] = 1;
  for (long long i = head[x]; i; i = e[i].nt) {
    if (e[i].v == f) continue;
    dfs1(e[i].v, x);
    s[x] += s[e[i].v];
    if (s[e[i].v] > s[son[x]]) {
      son[x] = e[i].v;
    }
  }
}
void add(long long x, long long f, long long v) {
  p[c[x]] += v;
  if (p[c[x]] > mx)
    mx = p[c[x]], sum = c[x];
  else if (p[c[x]] == mx)
    sum += c[x];
  for (long long i = head[x]; i; i = e[i].nt) {
    if (e[i].v != f && e[i].v != sn) add(e[i].v, x, v);
  }
}
void dfs2(long long x, long long f, long long t) {
  for (long long i = head[x]; i; i = e[i].nt) {
    if (e[i].v == f || e[i].v == son[x]) continue;
    dfs2(e[i].v, x, 0);
  }
  if (son[x]) dfs2(son[x], x, 1);
  sn = son[x];
  add(x, f, 1);
  sn = 0;
  ans[x] = sum;
  if (!t) add(x, f, -1), sum = mx = 0;
}
signed main() {
  n = read();
  for (long long i = 1; i <= n; i++) c[i] = read();
  for (long long i = 1; i <= n - 1; i++) {
    long long x = read(), y = read();
    add(x, y);
    add(y, x);
  }
  dfs1(1, 0);
  dfs2(1, 0, 0);
  for (long long i = 1; i <= n; i++) {
    cout << ans[i] << " ";
  }
}
