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;
template <int N, int M, class T>
struct Link {
  int HEAD[N], NXT[M], tot;
  T W[M];
  void clear() {
    memset(HEAD, 0, sizeof HEAD);
    tot = 0;
  }
  void add(int x, T w) {
    NXT[++tot] = HEAD[x];
    W[HEAD[x] = tot] = w;
  }
  T& operator[](int x) { return W[x]; }
};
Link<100005, 100005 << 1, int> G;
int siz[100005], hs[100005], c[100005], dfn[100005], en[100005], No, mp[100005];
void dfs(int o, int f) {
  if (!f) No = 0;
  dfn[o] = ++No;
  mp[No] = o;
  siz[o] = 1;
  hs[o] = 0;
  for (int k = G.HEAD[o]; k; k = G.NXT[k])
    if (G[k] != f) {
      dfs(G[k], o);
      siz[o] += siz[G[k]];
      if (siz[G[k]] > siz[hs[o]]) hs[o] = G[k];
    }
  en[o] = No;
}
long long sum[100005];
int cnt[100005], Max;
void add(int o, int d) {
  sum[cnt[c[o]]] -= c[o];
  sum[cnt[c[o]] += d] += c[o];
  Max = max(Max, cnt[c[o]]);
  if (!sum[Max]) Max--;
}
void upd(int o, int d) {
  for (int i = dfn[o]; i <= en[o]; i++) add(mp[i], d);
}
long long ans[100005];
void dsu(int o, int f, bool kep) {
  for (int k = G.HEAD[o]; k; k = G.NXT[k])
    if (G[k] != f && G[k] != hs[o]) dsu(G[k], o, 0);
  if (hs[o]) dsu(hs[o], o, 1);
  for (int k = G.HEAD[o]; k; k = G.NXT[k])
    if (G[k] != f && G[k] != hs[o]) upd(G[k], 1);
  add(o, 1);
  ans[o] = sum[Max];
  if (!kep) upd(o, -1);
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", c + i);
  for (int i = 1, a, b; i < n; i++) {
    scanf("%d%d", &a, &b);
    G.add(a, b);
    G.add(b, a);
  }
  dfs(1, 0);
  dsu(1, 0, 1);
  for (int i = 1; i <= n; i++) printf("%lld ", ans[i]);
  return 0;
}
