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 + 100;
int col[maxn], sz[maxn], son[maxn], cnt[maxn];
int n, big, maxx;
long long ans[maxn], sum;
vector<int> G[maxn];
void get_son(int u, int p) {
  sz[u] = 1;
  int len = G[u].size();
  for (int i = 0; i < len; i++) {
    int v = G[u][i];
    if (v != p) {
      get_son(v, u);
      sz[u] += sz[v];
      if (sz[v] > sz[son[u]]) son[u] = v;
    }
  }
}
void add(int u, int p, int x) {
  cnt[col[u]] += x;
  if (cnt[col[u]] > maxx) {
    sum = col[u];
    maxx = cnt[col[u]];
  } else if (cnt[col[u]] == maxx) {
    sum += col[u];
  }
  int len = G[u].size();
  for (int i = 0; i < len; i++) {
    int v = G[u][i];
    if (v != p && v != big) add(v, u, x);
  }
}
void dfs(int u, int p, bool keep) {
  int len = G[u].size();
  for (int i = 0; i < len; i++) {
    int v = G[u][i];
    if (v != p && v != son[u]) dfs(v, u, 0);
  }
  if (son[u]) {
    dfs(son[u], u, 1);
    big = son[u];
  }
  add(u, p, 1);
  big = 0;
  ans[u] = sum;
  if (!keep) {
    add(u, p, -1);
    maxx = 0;
    sum = 0;
  }
}
int main() {
  while (~scanf("%d", &n)) {
    big = 0;
    maxx = 0;
    sum = 0;
    for (int i = 1; i <= n; i++) {
      G[i].clear();
      sz[i] = son[i] = cnt[i] = ans[i] = 0;
    }
    for (int i = 1; i <= n; i++) scanf("%d", &col[i]);
    int u, v;
    for (int i = 0; i < n - 1; i++) {
      scanf("%d%d", &u, &v);
      G[u].push_back(v);
      G[v].push_back(u);
    }
    get_son(1, 0);
    dfs(1, 0, 1);
    for (int i = 1; i <= n; i++) {
      if (i != 1) printf(" ");
      printf("%lld", ans[i]);
    }
    printf("\n");
  }
  return 0;
}
