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;
const int maxn = 2e5 + 10;
bool vis[maxn];
int father[maxn], siz[maxn], son[maxn], color[maxn], n, tot, pre[maxn],
    last[maxn], other[maxn], x, y, cnt[maxn], num;
long long ans[maxn], temp;
void add(int x, int y) {
  pre[++tot] = last[x];
  last[x] = tot;
  other[tot] = y;
}
void dfs1(int x, int fa) {
  int p = last[x];
  siz[x] = 1;
  father[x] = fa;
  while (p) {
    if (other[p] != fa) {
      dfs1(other[p], x);
      siz[x] += siz[other[p]];
      if (siz[son[x]] < siz[other[p]]) son[x] = other[p];
    }
    p = pre[p];
  }
}
void calc(int x, int opt) {
  cnt[color[x]] += opt;
  if (opt > 0) {
    if (cnt[color[x]] == num) temp += color[x];
    if (cnt[color[x]] > num) {
      num = cnt[color[x]];
      temp = color[x];
    }
  }
  int p = last[x];
  while (p) {
    if (!vis[other[p]] && other[p] != father[x]) calc(other[p], opt);
    p = pre[p];
  }
}
void dfs2(int x, int isson) {
  int p = last[x];
  while (p) {
    if (other[p] != father[x] && other[p] != son[x]) dfs2(other[p], 0);
    p = pre[p];
  }
  if (son[x]) {
    dfs2(son[x], 1);
    vis[son[x]] = true;
  }
  calc(x, 1);
  ans[x] = temp;
  if (son[x]) vis[son[x]] = false;
  if (!isson) {
    calc(x, -1);
    num = 0;
    temp = 0;
  }
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &color[i]);
  for (int i = 1; i < n; ++i) {
    scanf("%d%d", &x, &y);
    add(x, y);
    add(y, x);
  }
  dfs1(1, 0);
  dfs2(1, 0);
  for (int i = 1; i < n; ++i) printf("%lld ", ans[i]);
  printf("%lld\n", ans[n]);
  return 0;
}
