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 x = 0;
  char ch = getchar();
  for (; ch < '0' || ch > '9'; ch = getchar())
    ;
  for (; ch <= '9' && ch >= '0'; x = x * 10 + ch - 48, ch = getchar())
    ;
  return x;
}
map<long long, long long> s[100010];
map<long long, long long>::iterator it;
long long i, j, k, m, n, x, y, w[100010], cnt, last[100010], a[100010],
    ans[100010], mx[100010], id[100010], as[100010];
struct edge {
  long long to, next;
} e[100010 << 1];
inline void add(long long u, long long v) {
  e[++cnt] = (edge){v, last[u]}, last[u] = cnt;
  e[++cnt] = (edge){u, last[v]}, last[v] = cnt;
}
inline void merge(long long x, long long y) {
  if (s[id[x]].size() < s[id[y]].size()) swap(id[x], id[y]);
  for (it = s[id[y]].begin(); it != s[id[y]].end(); ++it) {
    long long k = (*it).first, z = (*it).second;
    if ((s[id[x]][k] += z) > mx[id[x]]) mx[id[x]] = s[id[x]][k], ans[id[x]] = 0;
    if (s[id[x]][k] == mx[id[x]]) ans[id[x]] += k;
  }
}
void dfs(long long x, long long fa) {
  for (long long i = last[x], y; i; i = e[i].next)
    if ((y = e[i].to) != fa) dfs(y, x), merge(x, y);
  as[x] = ans[id[x]];
}
int main() {
  for (n = read(), i = 1; i <= n; ++i)
    s[i][ans[i] = w[i] = read()] = 1, mx[id[i] = i] = 1;
  for (i = 1; i < n; ++i) add(read(), read());
  for (dfs(1, 0), i = 1; i <= n; ++i) printf("%I64d ", as[i]);
}
