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;
const int maxn = 1e5 + 12;
int n, col[maxn], sz[maxn], maxx, cnt[maxn];
long long ans[maxn], can;
bool badboy[maxn];
vector<int> g[maxn];
void dastan(int v = 0, int p = -1) {
  sz[v] = 1;
  for (auto &u : g[v])
    if (u != p) dastan(u, v), sz[v] += sz[u];
}
void reval(int x) {
  if (maxx < ++cnt[x])
    maxx = cnt[x], can = x;
  else if (maxx == cnt[x])
    can += x;
}
void add(int v, int p = -1) {
  reval(col[v]);
  for (auto &u : g[v])
    if (u != p && !badboy[u]) add(u, v);
}
void rem(int v, int p = -1) {
  cnt[col[v]]--;
  for (auto &u : g[v])
    if (u != p && !badboy[u]) rem(u, v);
}
void dfs(int v = 0, int p = -1, bool hrh = 0) {
  int mx = 0, big = -1;
  for (auto &u : g[v])
    if (u != p && sz[u] > mx) mx = sz[u], big = u;
  for (auto &u : g[v])
    if (u != p && u != big) dfs(u, v, 1);
  if (big + 1) dfs(big, v), badboy[big] = 1;
  add(v, p);
  if (big + 1) badboy[big] = 0;
  ans[v] = can;
  if (hrh) rem(v, p), maxx = can = 0;
}
int main() {
  scanf("%d", &n);
  for_each(col, col + n, [](int &x) { scanf("%d", &x); });
  for (int i = 1, v, u; i < n; i++)
    scanf("%d%d", &v, &u), g[--v].push_back(--u), g[u].push_back(v);
  dastan(), dfs();
  for_each(ans, ans + n, [](long long &x) { printf("%lld ", x); });
  putchar('\n');
  return 0;
}
