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;
int n;
int a[1000005];
vector<int> v[1000005];
int root[1000005];
struct zz {
  int l, r, tot;
  long long sum;
};
int tot;
struct Tree {
  zz t[4000005];
  void clear() { memset(t, 0, sizeof t); }
  int Build_Tree(int p, int x, int l, int r) {
    if (!p) p = ++tot;
    if (l == r) {
      t[p].tot = 1;
      t[p].sum = x;
      return p;
    }
    int mid = (l + r) >> 1;
    if (x <= mid)
      t[p].l = Build_Tree(t[p].l, x, l, mid);
    else
      t[p].r = Build_Tree(t[p].r, x, mid + 1, r);
    if (t[t[p].l].tot == t[t[p].r].tot)
      t[p].sum = t[t[p].l].sum + t[t[p].r].sum;
    else {
      if (t[t[p].l].tot > t[t[p].r].tot)
        t[p].sum = t[t[p].l].sum;
      else
        t[p].sum = t[t[p].r].sum;
    }
    t[p].tot = max(t[t[p].l].tot, t[t[p].r].tot);
    return p;
  }
  int Change_Tree(int p, int q, int l, int r) {
    if (!q) return p;
    if (!p) return q;
    if (l == r) {
      t[p].tot += t[q].tot;
      t[p].sum = t[q].sum;
      return p;
    }
    int mid = (l + r) >> 1;
    t[p].l = Change_Tree(t[p].l, t[q].l, l, mid);
    t[p].r = Change_Tree(t[p].r, t[q].r, mid + 1, r);
    if (t[t[p].l].tot == t[t[p].r].tot)
      t[p].sum = t[t[p].l].sum + t[t[p].r].sum;
    else {
      if (t[t[p].l].tot > t[t[p].r].tot)
        t[p].sum = t[t[p].l].sum;
      else
        t[p].sum = t[t[p].r].sum;
    }
    t[p].tot = max(t[t[p].l].tot, t[t[p].r].tot);
    return p;
  }
  void Check(int x, int y) {
    printf("Start Check %d %d\n", x, y);
    for (int i = 1; i <= n; i++)
      printf("id:%d %lld %lld\n", i, t[root[i]].sum, t[root[i]].tot);
    printf("End\n\n");
  }
} t;
bool f[1000005];
long long ans[1000005];
void DFS(int x) {
  for (int i = 0; i < v[x].size(); i++) {
    int y = v[x][i];
    if (f[y]) continue;
    f[y] = 1;
    DFS(y);
    root[x] = t.Change_Tree(root[x], root[y], 1, n);
  }
  ans[x] = t.t[root[x]].sum;
}
int main() {
  t.clear();
  tot = 0;
  cin >> n;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &a[i]);
    root[i] = t.Build_Tree(root[i], a[i], 1, n);
  }
  for (int i = 1; i <= n - 1; i++) {
    int x, y;
    scanf("%d%d", &x, &y);
    v[x].push_back(y);
    v[y].push_back(x);
  }
  f[1] = 1;
  DFS(1);
  for (int i = 1; i <= n; i++) printf("%lld ", ans[i]);
  return 0;
}
