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 N = 150005;
struct Tree {
  vector<int> adj[N];
  int ld[N], rd[N], sz[N], id[N];
  int big[N];
  int n, ind, rt;
  void trprint() {
    cout << "rt=" << rt << ' ' << "n=" << n << endl;
    for (int i = 1; i <= n; ++i)
      cout << "i=" << i << ' ' << ld[i] << ' ' << rd[i] << ' ' << sz[i] << ' '
           << big[i] << endl;
  }
  void trinit() {
    for (int i = 1; i <= n; ++i) adj[i].clear();
    n = ind = 0;
    rt = 1;
  }
  inline void addedge(const int x, const int y) {
    adj[x].push_back(y);
    adj[y].push_back(x);
  }
  void trdfs(int x, int pre) {
    ld[x] = ++ind;
    id[ind] = x;
    sz[x] = 1;
    big[x] = 0;
    for (auto &y : adj[x])
      if (y != pre) {
        trdfs(y, x);
        sz[x] += sz[y];
        if (sz[y] > sz[big[x]]) big[x] = y;
      }
    rd[x] = ld[x] + sz[x] - 1;
  }
};
int cnt[N], mxt;
struct DsuOnTree : Tree {
  int col[N];
  long long ans[N];
  void dsuprint() {
    for (int i = 1; i <= n; ++i)
      cout << "i=" << i << ' ' << col[i] << ' ' << ans[i] << endl;
  }
  inline void add(const int x, const int t) { cnt[col[x]] += t; }
  inline void calcans(const int v, const int u) {
    if ((cnt[col[v]] + 1) == mxt)
      ans[u] += col[v];
    else if ((cnt[col[v]] + 1) > mxt)
      mxt = cnt[col[v]] + 1, ans[u] = col[v];
  }
  void relax(const int x) {
    mxt = 0;
    for (int i = ld[x]; i <= rd[x]; ++i) add(id[i], -1);
  }
  void dfs(int x, int pre, int keep) {
    for (auto &y : adj[x])
      if (y != pre && y != big[x]) dfs(y, x, 0);
    if (big[x]) dfs(big[x], x, 1);
    ans[x] = ans[big[x]];
    calcans(x, x);
    add(x, 1);
    for (auto &y : adj[x])
      if (y != pre && y != big[x])
        for (int i = ld[y]; i <= rd[y]; ++i) calcans(id[i], x), add(id[i], 1);
    if (!keep) relax(x);
  }
} tr;
int n;
int main() {
  tr.trinit();
  cin >> n;
  tr.n = n;
  for (int i = 1; i <= n; ++i) scanf("%d", &tr.col[i]);
  for (int i = 1, x, y; i < n; ++i) scanf("%d%d", &x, &y), tr.addedge(x, y);
  tr.trdfs(1, 0);
  tr.dfs(1, 0, 0);
  for (int i = 1; i <= n; ++i) printf("%lld ", tr.ans[i]);
  puts("");
  return 0;
}
