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>
#pragma GCC optimize(2)
using namespace std;
const int N = 1e6 + 5;
const int M = 1e4 + 5;
const int inf = 0x3f3f3f3f;
const long long mod = 1e9 + 7;
const double eps = 1e-9;
const long double pi = acos(-1.0L);
long long read() {
  long long x = 0, t = 1;
  char ch;
  while (!isdigit(ch = getchar()))
    if (ch == '-') t = -1;
  while (isdigit(ch)) {
    x = 10 * x + ch - '0';
    ch = getchar();
  }
  return x * t;
}
long long res, ma, ans[N];
int cnt[N], son[N], col[N], a[N];
vector<int> e[N];
void cal(int u, int pre, int x, int flag) {
  col[a[u]] += x;
  if (col[a[u]] > ma) {
    ma = col[a[u]];
    res = a[u];
  } else if (col[a[u]] == ma)
    res += a[u];
  for (auto v : e[u]) {
    if (v == pre || v == flag) continue;
    cal(v, u, x, flag);
  }
}
void dfs(int u, int pre) {
  cnt[u] = 1;
  for (auto v : e[u]) {
    if (v == pre) continue;
    dfs(v, u);
    cnt[u] += cnt[v];
    if (cnt[son[u]] < cnt[v]) son[u] = v;
  }
}
void dfs2(int u, int pre, int del) {
  for (auto v : e[u])
    if (v != pre && v != son[u]) dfs2(v, u, 0);
  if (son[u]) dfs2(son[u], u, 1);
  cal(u, pre, 1, son[u]);
  ans[u] = res;
  if (del == 0) cal(u, pre, -1, 0), res = 0, ma = 0;
}
int main() {
  int n = read();
  for (int i = 1; i <= n; i++) a[i] = read();
  for (int i = 1; i < n; i++) {
    int x = read(), y = read();
    e[x].push_back(y);
    e[y].push_back(x);
  }
  dfs(1, 0);
  dfs2(1, 0, 0);
  for (int i = 1; i <= n; i++) printf("%lld%c", ans[i], i == n ? '\n' : ' ');
  return 0;
}
