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>
template <class T>
void rd(T &x) {
  x = 0;
  T f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -1;
    ch = getchar();
  }
  while (ch <= '9' && ch >= '0') x = x * 10 + ch - '0', ch = getchar();
  x *= f;
}
using namespace std;
const int N = 1e5 + 10;
int n, c[N];
int fi[N], nxt[N << 1], to[N << 1], tot;
void link(int x, int y) {
  nxt[++tot] = fi[x];
  fi[x] = tot;
  to[tot] = y;
}
int siz[N];
void init(int x, int p) {
  siz[x] = 1;
  for (int i = fi[x]; i; i = nxt[i])
    if (to[i] != p) {
      init(to[i], x);
      siz[x] += siz[to[i]];
    }
}
long long ans[N];
int mx[N];
int cnt[N];
bool big[N];
long long tmp_ans;
int tmp_mx;
void add(int x, int p, int y) {
  cnt[c[x]] += y;
  if (cnt[c[x]] == tmp_mx)
    tmp_ans += c[x];
  else if (cnt[c[x]] > tmp_mx) {
    tmp_mx = cnt[c[x]];
    tmp_ans = c[x];
  }
  for (int i = fi[x]; i; i = nxt[i])
    if (to[i] != p && !big[to[i]]) add(to[i], x, y);
}
void dfs(int x, int p, int keep) {
  int son = 0;
  for (int i = fi[x]; i; i = nxt[i])
    if (to[i] != p && siz[to[i]] > siz[son]) son = to[i];
  for (int i = fi[x]; i; i = nxt[i])
    if (to[i] != p && to[i] != son) dfs(to[i], x, 0);
  if (son) dfs(son, x, 1), big[son] = 1;
  tmp_ans = ans[son];
  tmp_mx = mx[son];
  add(x, p, 1);
  ans[x] = tmp_ans;
  mx[x] = tmp_mx;
  if (son) big[son] = 0;
  if (!keep) add(x, p, -1);
}
int main() {
  rd(n);
  for (int i = 1; i <= n; i++) rd(c[i]);
  for (int i = 1; i < n; i++) {
    int x, y;
    rd(x);
    rd(y);
    link(x, y);
    link(y, x);
  }
  init(1, 0);
  dfs(1, 0, 0);
  for (int i = 1; i <= n; i++) printf("%lld%c", ans[i], i == n ? '\n' : ' ');
  return 0;
}
