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;
inline void read(int& x) {
  static char c;
  while (!isdigit(c = getchar()))
    ;
  x = c ^ 48;
  while (isdigit(c = getchar())) x = (x * 10) + (c ^ 48);
}
const int N = 100005;
int n, Son, u, v;
int Col[N], Buc[N];
long long Ans[N], Mx, Sum;
struct Edge {
  int v, nxt;
} ed[N << 1];
int head[N], cnt;
inline void adde(int u, int v) {
  ed[++cnt] = (Edge){v, head[u]}, head[u] = cnt;
}
inline void add(int u, int v) { adde(u, v), adde(v, u); }
int siz[N], son[N];
void P_Son(int now, int fa) {
  siz[now] = 1;
  for (register int i = head[now]; i; i = ed[i].nxt) {
    int v = ed[i].v;
    if (v == fa) continue;
    P_Son(v, now), siz[now] += siz[v];
    if (siz[v] > siz[son[now]]) son[now] = v;
  }
}
void Modify(int now, int fa, int B) {
  Buc[Col[now]] += B;
  if (Buc[Col[now]] > Mx)
    Mx = Buc[Col[now]], Sum = Col[now];
  else if (Buc[Col[now]] == Mx)
    Sum += Col[now];
  for (register int i = head[now]; i; i = ed[i].nxt) {
    int v = ed[i].v;
    if ((v ^ fa) && (v ^ Son)) Modify(v, now, B);
  }
}
void Calc(int now, int fa, bool B) {
  for (register int i = head[now]; i; i = ed[i].nxt) {
    int v = ed[i].v;
    if ((v ^ fa) && (v ^ son[now])) Calc(v, now, 0);
  }
  if (son[now]) Calc(son[now], now, 1), Son = son[now];
  Modify(now, fa, 1), Son = 0;
  Ans[now] = Sum;
  if (!B) Modify(now, fa, -1), Sum = 0, Mx = 0;
}
inline void Work() {
  read(n);
  for (register int i = 1; i <= n; ++i) read(Col[i]);
  for (register int i = 1; i < n; ++i) read(u), read(v), add(u, v);
  P_Son(1, 1);
  Calc(1, 1, 1);
  for (register int i = 1; i <= n; ++i) printf("%lld ", Ans[i]);
}
int main() { Work(); }
