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;
using ll = long long;
const int N = 1e5 + 1;
void read(int& n) {
  n = 0;
  while (1) {
    char c = getchar();
    if (c == ' ' || c == '\n') break;
    n = (n << 1) + (n << 3) + c - '0';
  }
}
void read(ll& n) {
  n = 0;
  while (1) {
    char c = getchar();
    if (c == ' ' || c == '\n') break;
    n = (n << 1) + (n << 3) + c - '0';
  }
}
void write(int n) {
  if (n > 9)
    write(n / 10), putchar(n % 10 + '0');
  else
    putchar(n + '0');
}
void write(ll n) {
  if (n > 9)
    write(n / 10), putchar(n % 10 + '0');
  else
    putchar(n + '0');
}
template <class F, class... Ts>
void read(F& f, Ts&... ts) {
  read(f);
  read(ts...);
}
template <class F, class... Ts>
void write(F& f, Ts&... ts) {
  write(f);
  putchar(' ');
  write(ts...);
}
template <class... Ts>
void writeln(const Ts&... ts) {
  write(ts...);
  putchar('\n');
}
int n, sz[N], col[N], cnt[N];
ll ans[N], res[N];
vector<int> adj[N], vec[N];
void dfs_size(int v, int p = 0) {
  sz[v] = 1;
  for (auto u : adj[v]) {
    if (u != p) {
      dfs_size(u, v);
      sz[v] += sz[u];
    }
  }
}
void update(int& mx, int v, int x) {
  ans[cnt[col[v]]] -= col[v];
  cnt[col[v]] += x;
  ans[cnt[col[v]]] += col[v];
  mx = max(mx, cnt[col[v]]);
}
int dfs(int v, int p = 0, bool keep = 0) {
  int mx = 0, Max = -1, bigchild = -1;
  for (auto u : adj[v]) {
    if (u != p && Max < sz[u]) {
      Max = sz[u];
      bigchild = u;
    }
  }
  for (auto u : adj[v]) {
    if (u != p && u != bigchild) {
      dfs(u, v, 0);
    }
  }
  if (bigchild != -1) {
    mx = max(mx, dfs(bigchild, v, 1));
    vec[bigchild].swap(vec[v]);
  }
  vec[v].push_back(v);
  update(mx, v, 1);
  for (auto u : adj[v]) {
    if (u != p && u != bigchild) {
      for (auto x : vec[u]) {
        update(mx, x, 1);
        vec[v].push_back(x);
      }
    }
  }
  res[v] = ans[mx];
  if (keep == 0) {
    for (auto u : vec[v]) {
      update(mx, u, -1);
    }
  }
  return mx;
}
int main() {
  read(n);
  for (int i = 1; i <= n; i++) {
    read(col[i]);
  }
  for (int i = 1, u, v; i < n; i++) {
    read(u, v);
    adj[u].push_back(v);
    adj[v].push_back(u);
  }
  dfs_size(1), dfs(1);
  for (int i = 1; i <= n; i++) {
    write(res[i]);
    putchar(' ');
  }
}
