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;
struct node {
  int to, next;
} e[100005 << 1];
int head[100005], val[100005], col[100005], maxx, siz[100005], son[100005], n,
    cnt;
long long ans[100005], sum;
void add(int x, int y) {
  e[cnt] = (node){y, head[x]};
  head[x] = cnt++;
}
void dfs1(int x, int from) {
  siz[x] = 1;
  for (int i = head[x]; i != -1; i = e[i].next) {
    int to1 = e[i].to;
    if (to1 != from)
      dfs1(to1, x), siz[x] += siz[to1],
          son[x] = (siz[to1] > siz[son[x]] ? to1 : son[x]);
  }
}
void add(int x, int from, int c) {
  col[val[x]] += c;
  if (col[val[x]] >= maxx && c > 0) {
    if (col[val[x]] > maxx) maxx = col[val[x]], sum = 0;
    sum += val[x];
  }
  for (int i = head[x]; i != -1; i = e[i].next) {
    int to1 = e[i].to;
    if (to1 != from) add(to1, x, c);
  }
}
void dfs2(int x, int from, int op) {
  for (int i = head[x]; i != -1; i = e[i].next) {
    int to1 = e[i].to;
    if (to1 != from && to1 != son[x]) dfs2(to1, x, 0);
  }
  if (son[x]) dfs2(son[x], x, 1);
  for (int i = head[x]; i != -1; i = e[i].next) {
    int to1 = e[i].to;
    if (to1 != from && to1 != son[x]) add(to1, x, 1);
  }
  col[val[x]]++;
  if (col[val[x]] >= maxx) {
    if (col[val[x]] > maxx) maxx = col[val[x]], sum = 0;
    sum += val[x];
  }
  ans[x] = sum;
  if (!op) add(x, from, -1), sum = 0, maxx = 0;
}
int main() {
  scanf("%d", &n);
  memset(head, -1, sizeof(head));
  for (int i = 1; i <= n; i++) scanf("%d", &val[i]);
  for (int i = 1, x, y; i < n; i++) scanf("%d%d", &x, &y), add(x, y), add(y, x);
  dfs1(1, 0);
  dfs2(1, 0, 1);
  for (int i = 1; i <= n; i++) printf("%lld ", ans[i]);
  return 0;
}
