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;
int n;
int c[maxn];
int ecnt;
int head[maxn], nxt[maxn << 1], to[maxn << 1];
map<int, int> cnt[maxn];
map<int, long long> sum[maxn];
long long ans[maxn];
inline void addedge(int u, int v) {
  to[++ecnt] = v;
  nxt[ecnt] = head[u];
  head[u] = ecnt;
}
void dfs(int u, int pre) {
  cnt[u][c[u]] = 1;
  sum[u][1] = c[u];
  for (int i = head[u]; i; i = nxt[i]) {
    int v = to[i];
    if (v == pre) continue;
    dfs(v, u);
    if (cnt[u].size() < cnt[v].size()) {
      swap(cnt[u], cnt[v]);
      swap(sum[u], sum[v]);
    }
    for (map<int, int>::iterator it = cnt[v].begin(); it != cnt[v].end();
         ++it) {
      cnt[u][it->first] += it->second;
      sum[u][cnt[u][it->first]] += it->first;
    }
    cnt[v].clear();
    sum[v].clear();
  }
  ans[u] = sum[u].rbegin()->second;
}
int main() {
  while (~scanf("%d", &n)) {
    ecnt = 0;
    cnt[1].clear();
    sum[1].clear();
    memset(head, 0, sizeof(head));
    for (int i = 1; i <= n; ++i) {
      scanf("%d", &c[i]);
    }
    for (int i = 1; i < n; ++i) {
      int x, y;
      scanf("%d%d", &x, &y);
      addedge(x, y);
      addedge(y, x);
    }
    dfs(1, -1);
    for (int i = 1; i <= n; ++i) {
      printf("%lld%c", ans[i], i == n ? '\n' : ' ');
    }
  }
  return 0;
}
