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 = 1e5 + 10;
inline int read() {
  char c = getchar();
  int x = 0, f = 1;
  while (c < '0' || c > '9') {
    if (c == '-') f = -1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
  return x * f;
}
int N, col[MAXN], son[MAXN], siz[MAXN], cnt[MAXN], Mx, Son;
long long sum = 0, ans[MAXN];
vector<int> v[MAXN];
void dfs(int x, int fa) {
  siz[x] = 1;
  for (int i = 0; i < v[x].size(); i++) {
    int to = v[x][i];
    if (to == fa) continue;
    dfs(to, x);
    siz[x] += siz[to];
    if (siz[to] > siz[son[x]]) son[x] = to;
  }
}
void add(int x, int fa, int val) {
  for (int i = 0; i < v[x].size(); i++) {
    int to = v[x][i];
    if (to == fa || to == Son) continue;
    add(to, x, val);
  }
  cnt[col[x]] += val;
  if (cnt[col[x]] > Mx)
    Mx = cnt[col[x]], sum = col[x];
  else if (cnt[col[x]] == Mx)
    sum += (long long)col[x];
}
void dfs2(int x, int fa, int opt) {
  for (int i = 0; i < v[x].size(); i++) {
    int to = v[x][i];
    if (to == fa) continue;
    if (to != son[x]) dfs2(to, x, 0);
  }
  if (son[x]) dfs2(son[x], x, 1), Son = son[x];
  add(x, fa, 1);
  Son = 0;
  ans[x] = sum;
  if (!opt) add(x, fa, -1), sum = 0, Mx = 0;
}
int main() {
  N = read();
  for (int i = 1; i <= N; i++) col[i] = read();
  for (int i = 1; i <= N - 1; i++) {
    int x = read(), y = read();
    v[x].push_back(y);
    v[y].push_back(x);
  }
  dfs(1, 0);
  dfs2(1, 0, 0);
  for (int i = 1; i <= N; i++) printf("%I64d ", ans[i]);
  return 0;
}
