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>
const int N = 100001;
int n, a[N], sz[N], hs[N], c[N], mx;
bool vis[N];
long long ans[N], s;
std::vector<int> v[N];
void change(int x, int fa, int k) {
  if (k > 0)
    c[a[x]]++;
  else
    c[a[x]] = 0;
  if (k > 0 && c[a[x]] >= mx) {
    if (c[a[x]] > mx) mx = c[a[x]], s = 0;
    s += a[x];
  }
  for (auto y : v[x])
    if (y != fa && !vis[y]) change(y, x, k);
}
void dfs1(int x, int fa = 0) {
  sz[x] = 1;
  for (auto y : v[x])
    if (y != fa) {
      dfs1(y, x);
      sz[x] += sz[y];
      if (sz[y] > sz[hs[x]]) hs[x] = y;
    }
}
void dfs2(int x, int fa = 0, bool b = false) {
  for (auto y : v[x])
    if (y != fa && y != hs[x]) dfs2(y, x);
  if (hs[x]) dfs2(hs[x], x, true), vis[hs[x]] = true;
  change(x, fa, 1);
  ans[x] = s;
  vis[hs[x]] = false;
  if (!b) change(x, fa, -1), mx = s = 0;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (int x, y, i = 1; i < n; ++i) {
    scanf("%d%d", &x, &y);
    v[x].push_back(y);
    v[y].push_back(x);
  }
  dfs1(1);
  dfs2(1);
  for (int i = 1; i <= n; ++i) printf("%lld ", ans[i]);
}
