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 = 2e5 + 8;
using namespace std;
int n, a[N], si[N];
int tot, now[N], to[N << 1], pre[N << 1];
long long mx, sum, ans[N], c[N];
bool bson[N];
void add(int x, int y) {
  to[++tot] = y;
  pre[tot] = now[x];
  now[x] = tot;
}
void ycl(int u, int fa) {
  si[u]++;
  int mx = 0, ch = 0;
  for (int i = now[u]; i; i = pre[i])
    if (to[i] != fa) {
      ycl(to[i], u);
      si[u] += si[to[i]];
      if (si[to[i]] > mx) {
        mx = si[to[i]];
        ch = to[i];
      }
    }
  if (ch) bson[ch] = true;
}
void dfs(int u, int fa, int bs) {
  c[a[u]]++;
  if (c[a[u]] > mx) {
    mx = c[a[u]];
    sum = a[u];
  } else if (c[a[u]] == mx)
    sum += a[u];
  for (int i = now[u]; i; i = pre[i])
    if (to[i] != fa && to[i] != bs) dfs(to[i], u, bs);
}
inline void clear(int u, int fa) {
  c[a[u]]--;
  for (int i = now[u]; i; i = pre[i])
    if (to[i] != fa) clear(to[i], u);
}
inline void dsu(int u, int fa) {
  int ch = 0;
  for (int i = now[u]; i; i = pre[i])
    if (to[i] != fa) {
      if (bson[to[i]])
        ch = to[i];
      else {
        dsu(to[i], u);
        clear(to[i], u);
        mx = sum = 0;
      }
    }
  if (ch) dsu(ch, u);
  dfs(u, fa, ch);
  ans[u] = sum;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &a[i]);
  for (int i = 1, x, y; i < n; ++i) {
    scanf("%d%d", &x, &y);
    add(x, y), add(y, x);
  }
  ycl(1, 0);
  dsu(1, 0);
  for (int i = 1; i <= n; ++i) printf("%lld ", ans[i]);
  return 0;
}
