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;
inline long long rd() {
  long long s = 0;
  char c = getchar();
  bool f = 0;
  for (; c < '0' || c > '9'; c = getchar()) f |= c == '-';
  for (; c >= '0' && c <= '9'; c = getchar())
    s = (s << 3) + (s << 1) + (c ^ 48);
  return f ? ~s + 1 : s;
}
const int N = 1e5 + 10;
int n, a[N];
vector<int> e[N];
int sz[N], son[N], Son;
long long ans[N];
int cnt[N];
long long sum, maxcnt;
void dfs1(int x, int fa) {
  int mx = 0;
  sz[x] = 1;
  for (auto y : e[x]) {
    if (y == fa) continue;
    dfs1(y, x);
    sz[x] += sz[y];
    if (sz[y] > mx) {
      mx = sz[y];
      son[x] = y;
    }
  }
}
void add(int x, int fa) {
  ++cnt[a[x]];
  if (cnt[a[x]] > maxcnt) {
    maxcnt = cnt[a[x]];
    sum = a[x];
  } else if (cnt[a[x]] == maxcnt)
    sum += a[x];
  for (auto y : e[x]) {
    if (y == fa || y == Son) continue;
    add(y, x);
  }
}
void del(int x, int fa) {
  --cnt[a[x]];
  for (auto y : e[x]) {
    if (y == fa) continue;
    del(y, x);
  }
}
void dfs2(int x, int fa, bool f) {
  for (auto y : e[x]) {
    if (y == fa || y == son[x]) continue;
    dfs2(y, x, 1);
  }
  if (son[x]) dfs2(son[x], x, 0);
  Son = son[x];
  add(x, fa);
  ans[x] = sum;
  if (f) {
    del(x, fa);
    sum = 0;
    maxcnt = 0;
  }
}
int main() {
  n = rd();
  for (int i = 1; i <= n; ++i) a[i] = rd();
  for (int i = 1; i < n; ++i) {
    int x = rd(), y = rd();
    e[x].push_back(y);
    e[y].push_back(x);
  }
  dfs1(1, 0);
  dfs2(1, 0, 0);
  for (int i = 1; i <= n; ++i) printf("%lld ", ans[i]);
}
