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 long long MXN = 3e5 + 1;
const long long MAXN = 1e3 + 1;
const long long MOD = 1e9 + 7;
const long long INF = 1e18;
inline void f_math(bool ok = 0) {
  ios_base::sync_with_stdio(0);
  cin.tie(0);
  cout.tie(0);
  if (ok) {
    freopen("aquarium.in", "r", stdin);
    freopen("aquarium.out", "w", stdout);
    ;
  }
}
long long n, sum[MXN], sz[MXN], mx[MXN], num[MXN], ans[MXN], c[MXN];
vector<int> g[MXN];
map<int, int> cnt[MXN];
void go(int v, int c, int s) {
  cnt[v][c] += s;
  if (cnt[v][c] > mx[v]) {
    sum[v] = c;
    mx[v] = cnt[v][c];
  } else if (cnt[v][c] == mx[v]) {
    sum[v] += c;
  }
}
void dfs(int v, int pr = -1) {
  int res = -1;
  sz[v] = 1;
  for (auto to : g[v]) {
    if (to == pr) continue;
    dfs(to, v);
    sz[v] += sz[to];
    if (res == -1 || sz[to] > sz[res]) res = to;
  }
  if (res != -1) {
    int w = res;
    res = num[res];
    num[v] = res;
    go(res, c[v], 1);
    for (auto to : g[v]) {
      if (to != pr && to != w) {
        for (auto c : cnt[num[to]]) {
          go(res, c.first, c.second);
        }
      }
    }
  } else {
    num[v] = v;
    go(v, c[v], 1);
  }
  ans[v] = sum[num[v]];
}
int main() {
  f_math();
  cin >> n;
  for (int i = 1; i <= n; i++) {
    cin >> c[i];
  }
  for (int i = 1; i < n; i++) {
    int u, v;
    cin >> u >> v;
    g[u].push_back(v);
    g[v].push_back(u);
  }
  dfs(1);
  for (int i = 1; i <= n; i++) {
    cout << ans[i] << ' ';
  }
  return 0;
}
