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>
struct edge {
  int to, nxt;
} e[100010 << 1];
int head[100010], a[100010], w[100010], st[100010], size[100010], son[100010],
    top = 0, mx = 0, sz = 0, n, i, x, y;
long long ans[100010], sum = 0;
int getnum() {
  int s = 0;
  char c = getchar();
  for (; !isdigit(c); c = getchar())
    ;
  for (; isdigit(c); c = getchar()) s = s * 10 + c - '0';
  return s;
}
void addedge(int x, int y) {
  e[++sz].to = y;
  e[sz].nxt = head[x];
  head[x] = sz;
}
void clear() {
  for (; top; top--) w[st[top]] = 0;
  mx = sum = 0;
}
void add(int x) {
  if (!w[a[x]]) st[++top] = a[x];
  w[a[x]]++;
  if (w[a[x]] > mx) {
    mx = w[a[x]];
    sum = a[x];
  } else if (w[a[x]] == mx)
    sum += 1LL * a[x];
}
void dfs1(int x, int y) {
  int i;
  size[x] = 1;
  for (i = head[x]; i; i = e[i].nxt)
    if (e[i].to != y) {
      dfs1(e[i].to, x);
      size[x] += size[e[i].to];
      if (size[e[i].to] > size[son[x]]) son[x] = e[i].to;
    }
}
void work(int x, int y) {
  int i;
  add(x);
  for (i = head[x]; i; i = e[i].nxt)
    if (e[i].to != y) work(e[i].to, x);
}
void dfs2(int x, int y) {
  int i;
  for (i = head[x]; i; i = e[i].nxt)
    if ((e[i].to != y) && (e[i].to != son[x])) {
      dfs2(e[i].to, x);
      clear();
    }
  if (son[x]) dfs2(son[x], x);
  for (i = head[x]; i; i = e[i].nxt)
    if ((e[i].to != y) && (e[i].to != son[x])) work(e[i].to, x);
  add(x);
  ans[x] = sum;
}
int main() {
  n = getnum();
  for (i = 1; i <= n; i++) a[i] = getnum();
  for (i = 1; i < n; i++) {
    x = getnum();
    y = getnum();
    addedge(x, y);
    addedge(y, x);
  }
  dfs1(1, 0);
  dfs2(1, 0);
  for (i = 1; i <= n; i++) printf("%I64d ", ans[i]);
  return 0;
}
