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>
const int maxn = 1e5 + 1000;
long long ans[maxn], tot;
int head[maxn], _cnt, n, mx;
int fa[maxn], siz[maxn], son[maxn], cnt[maxn], col[maxn], isson[maxn];
struct Edge {
  int to, next;
} edge[maxn << 1];
inline int max(int a, int b) { return a < b ? b : a; }
inline void add(int u, int v) {
  edge[++_cnt].next = head[u];
  edge[_cnt].to = v;
  head[u] = _cnt;
}
void dfs(int u, int f) {
  fa[u] = f;
  siz[u] = 1;
  int maxs = -1;
  for (int i = head[u]; i; i = edge[i].next) {
    int v = edge[i].to;
    if (v != f) {
      dfs(v, u);
      siz[u] += siz[v];
      if (siz[v] > maxs) maxs = siz[v], son[u] = v;
    }
  }
  if (son[u]) isson[son[u]] = true;
}
void AddAns(int u, int x, int Son) {
  if ((cnt[col[u]] += x) > mx)
    mx = cnt[col[u]], tot = col[u];
  else if (cnt[col[u]] == mx)
    tot += (long long)col[u];
  for (int i = head[u]; i; i = edge[i].next) {
    int v = edge[i].to;
    if (v != fa[u] && v != Son) AddAns(v, x, Son);
  }
}
void solve(int u, int f, int keep) {
  for (int i = head[u]; i; i = edge[i].next) {
    int v = edge[i].to;
    if (v != f && v != son[u]) solve(v, u, 0);
  }
  if (son[u]) solve(son[u], u, 1);
  AddAns(u, 1, son[u]);
  ans[u] = tot;
  if (!keep) AddAns(u, -1, 0), mx = tot = 0;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", col + i);
  for (int i = 1, u, v; i < n; ++i) {
    scanf("%d%d", &u, &v);
    add(u, v);
    add(v, u);
  }
  dfs(1, 1);
  solve(1, 0, 0);
  for (int i = 1; i <= n; ++i) printf("%I64d ", ans[i]);
}
