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 int read() {
  int nm = 0, fh = 1;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    if (c == '-') fh = -1;
  for (; isdigit(c); c = getchar()) nm = nm * 10 + c - '0';
  return nm * fh;
}
int n, tot, to[100020 << 1], nt[100020 << 1], fs[100020];
int col[100020], T[100020], sz[100020], fa[100020], mxs[100020], mx, dt;
long long ans[100020], sum;
inline void link(int u, int v) { to[++tot] = v, nt[tot] = fs[u], fs[u] = tot; }
inline void dfs(int u, int last = 0) {
  sz[u] = 1, fa[u] = last;
  for (int i = fs[u]; i; i = nt[i])
    if (to[i] ^ last) {
      dfs(to[i], u), sz[u] += sz[to[i]];
      if (sz[to[i]] > sz[mxs[u]]) mxs[u] = to[i];
    }
}
inline void chg(int u, int Lim) {
  T[col[u]] += dt;
  if (T[col[u]] > mx)
    mx = T[col[u]], sum = col[u];
  else if (T[col[u]] == mx)
    sum += (long long)col[u];
  for (int i = fs[u]; i; i = nt[i])
    if ((to[i] ^ fa[u]) && (to[i] ^ Lim)) chg(to[i], Lim);
}
inline void dfst(int u, int flag) {
  for (int i = fs[u]; i; i = nt[i])
    if ((to[i] ^ fa[u]) && (to[i] ^ mxs[u])) dfst(to[i], 1);
  if (mxs[u]) dfst(mxs[u], 0);
  dt = 1, chg(u, mxs[u]), ans[u] = sum;
  if (flag) dt = -1, chg(u, 0), mx = 0, sum = 0ll;
}
int main() {
  n = read();
  for (int i = 1; i <= n; i++) col[i] = read();
  for (int i = 1; i < n; i++) {
    int u = read(), v = read();
    link(u, v), link(v, u);
  }
  dfs(1), dfst(1, 0);
  for (int i = 1; i <= n; i++) printf("%lld ", ans[i]);
  puts("");
  return 0;
}
