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 long long LINF = 0x3f3f3f3f3f3f3f3f;
const long long NLINF = 0xf7f7f7f7f7f7f7f7;
const int INF = 0x3f3f3f3f, NINF = 0xf7f7f7f7;
const int MOD1 = 1e9 + 7, MOD2 = 1e9 + 9;
const int N = 1e5 + 10;
vector<int> adj[N];
int n, c[N], sz[N], tin[N], tout[N], lt[N], freq[N], cnt = 0, curmx = 0;
long long ans[N], val[N];
void dfs(int v, int p) {
  sz[v] = 1, tin[v] = ++cnt, lt[cnt] = v;
  for (auto &u : adj[v])
    if (u != p) dfs(u, v), sz[v] += sz[u];
  tout[v] = cnt;
}
void proc(int i, int type) {
  int curv = lt[i];
  if (freq[c[curv]]) val[freq[c[curv]]] -= c[curv];
  if (type == 1) {
    freq[c[curv]]++;
    if (freq[c[curv]] > curmx) curmx = freq[c[curv]];
    val[freq[c[curv]]] += c[curv];
  } else if (freq[c[curv]]) {
    freq[c[curv]]--;
    if (val[curmx] == 0ll) curmx--;
    val[freq[c[curv]]] += c[curv];
  }
}
void solve(int v, int p, bool heavy) {
  int mx = -1, big = -1;
  for (auto &u : adj[v])
    if (u != p && sz[u] > mx) mx = sz[u], big = u;
  for (auto &u : adj[v])
    if (u != p && u != big) solve(u, v, 0);
  if (big != -1) solve(big, v, 1);
  proc(tin[v], 1);
  for (auto &u : adj[v])
    if (u != p && u != big)
      for (int i = tin[u]; i <= tout[u]; i++) proc(i, 1);
  ans[v] = val[curmx];
  if (!heavy)
    for (int i = tin[v]; i <= tout[v]; i++) proc(i, 0);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &c[i]);
  for (int i = 1, u, v; i < n; i++) {
    scanf("%d%d", &u, &v);
    adj[u].emplace_back(v), adj[v].emplace_back(u);
  }
  dfs(1, 0);
  solve(1, 0, 0);
  for (int i = 1; i <= n; i++) printf("%lld%c", ans[i], " \n"[i == n]);
}
