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 = 100010, M = N << 1;
int n;
int w[N];
int head[N], nxt[M], to[M], tot;
long long sum[N], ans;
int cnt[N], mx, son[N], sz[N];
void add(int u, int v) {
  nxt[++tot] = head[u];
  head[u] = tot;
  to[tot] = v;
}
void dfs_son(int u, int fa) {
  sz[u] = 1;
  for (int i = head[u]; i; i = nxt[i]) {
    if (to[i] == fa) continue;
    dfs_son(to[i], u);
    sz[u] += sz[to[i]];
    if (sz[to[i]] > sz[son[u]]) son[u] = to[i];
  }
}
void update(int u, int fa, int x, int pson) {
  cnt[w[u]] += x;
  if (cnt[w[u]] > mx)
    mx = cnt[w[u]], ans = w[u];
  else if (cnt[w[u]] == mx)
    ans += w[u];
  for (int i = head[u]; i; i = nxt[i]) {
    if (to[i] == fa || to[i] == pson) continue;
    update(to[i], u, x, pson);
  }
}
void dfs(int u, int fa, int op) {
  for (int i = head[u]; i; i = nxt[i]) {
    if (to[i] == fa || to[i] == son[u]) continue;
    dfs(to[i], u, 0);
  }
  if (son[u]) dfs(son[u], u, 1);
  update(u, fa, 1, son[u]);
  sum[u] = ans;
  if (!op) update(u, fa, -1, 0), ans = mx = 0;
}
void solve() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &w[i]);
  int u, v;
  for (int i = 1; i < n; i++) {
    scanf("%d %d", &u, &v);
    add(u, v);
    add(v, u);
  }
  dfs_son(1, -1);
  dfs(1, -1, 1);
  for (int i = 1; i <= n; i++) printf("%lld ", sum[i]);
  puts("");
}
int main() {
  solve();
  return 0;
}
