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 = 1000010;
int n, u, v, cnt, tot[N], mx, col[N], sz[N], son[N], h[N], to[N], nxt[N];
long long ans[N], sm;
bool skip[N];
void add(int u, int v) {
  to[++cnt] = v;
  nxt[cnt] = h[u];
  h[u] = cnt;
}
void get(int x, int fa) {
  sz[x] = 1;
  for (int i = h[x], k; i; i = nxt[i])
    if ((k = to[i]) != fa) {
      get(k, x);
      sz[x] += sz[k];
      if (sz[k] > sz[son[x]]) son[x] = k;
    }
}
void dfs(int x, int fa, int op) {
  tot[col[x]] += op;
  if (op > 0 && tot[col[x]] >= mx) {
    if (tot[col[x]] > mx) sm = 0, mx = tot[col[x]];
    sm += col[x];
  }
  for (int i = h[x], k; i; i = nxt[i])
    if ((k = to[i]) != fa && !skip[k]) dfs(k, x, op);
}
void work(int x, int fa, bool cl) {
  for (int i = h[x], k; i; i = nxt[i])
    if ((k = to[i]) != fa && k != son[x]) work(k, x, 1);
  if (son[x]) work(son[x], x, 0), skip[son[x]] = 1;
  dfs(x, fa, 1);
  ans[x] = sm;
  skip[son[x]] = 0;
  if (cl) dfs(x, fa, -1), mx = sm = 0;
}
int main() {
  scanf("%d", &n);
  for (int i = (1); i <= (n); i++) scanf("%d", &col[i]);
  for (int i = (2); i <= (n); i++) scanf("%d%d", &u, &v), add(u, v), add(v, u);
  get(1, 0);
  work(1, 0, 0);
  for (int i = (1); i <= (n); i++) cout << ans[i] << ' ';
  return 0;
}
