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 = 100005;
int n, CNT[maxN], Best[maxN], sz[maxN], fa[maxN], dfn[maxN], stt[maxN],
    ext[maxN], tot = 0;
long long c[maxN], ans[maxN];
pair<long long, long long> mx = {0, 0};
vector<int> Nxt[maxN];
void dfs_reset(int u, int p) {
  sz[u] = 1;
  fa[u] = p;
  dfn[++tot] = u;
  stt[u] = tot;
  int B = 0;
  for (int i = 0; i < Nxt[u].size(); i++) {
    int v = Nxt[u][i];
    if (v == p) continue;
    dfs_reset(v, u);
    sz[u] += sz[v];
    if (sz[v] > B) {
      B = sz[v];
      Best[u] = v;
    }
  }
  ext[u] = tot;
}
void DSU_on_tree(int u, bool clr) {
  for (int i = 0; i < Nxt[u].size(); i++) {
    int v = Nxt[u][i];
    if (v == fa[u] || v == Best[u]) continue;
    DSU_on_tree(v, 1);
  }
  if (Best[u] != 0) {
    DSU_on_tree(Best[u], 0);
  }
  CNT[c[u]]++;
  if (CNT[c[u]] > mx.second) {
    mx = {c[u], CNT[c[u]]};
  } else if (CNT[c[u]] == mx.second) {
    mx.first += c[u];
  }
  for (int i = 0; i < Nxt[u].size(); i++) {
    int v = Nxt[u][i];
    if (v == fa[u] || v == Best[u]) continue;
    for (int j = stt[v]; j <= ext[v]; j++) {
      int k = dfn[j];
      CNT[c[k]]++;
      if (CNT[c[k]] > mx.second) {
        mx = {c[k], CNT[c[k]]};
      } else if (CNT[c[k]] == mx.second) {
        mx = {mx.first + c[k], CNT[c[k]]};
      }
    }
  }
  ans[u] = mx.first;
  if (clr) {
    for (int i = stt[u]; i <= ext[u]; i++) {
      CNT[c[dfn[i]]]--;
    }
    mx = {0, 0};
  }
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> c[i];
  }
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    Nxt[u].push_back(v);
    Nxt[v].push_back(u);
  }
  dfs_reset(1, 0);
  DSU_on_tree(1, 0);
  for (int i = 1; i <= n; i++) {
    cout << ans[i] << " ";
  }
  cout << endl;
}
