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 N = 1e5 + 9;
int n, c[N], head[N], tot;
struct pp {
  int nxt, to;
} g[N << 1];
void add(int u, int v) { g[++tot].nxt = head[u], g[tot].to = v, head[u] = tot; }
int siz[N], son[N];
void pre_dfs(int u, int fa) {
  siz[u] = 1, son[u] = 0;
  for (int i = head[u]; i != -1; i = g[i].nxt) {
    int v = g[i].to;
    if (v == fa) continue;
    pre_dfs(v, u);
    siz[u] += siz[v];
    if (siz[v] > siz[son[u]]) son[u] = v;
  }
}
int a[N];
long long mx, sum, ans[N];
void solve(int u, int fa) {
  a[c[u]]--;
  for (int i = head[u]; i != -1; i = g[i].nxt) {
    int v = g[i].to;
    if (v == fa) continue;
    solve(v, u);
  }
}
void DFS(int u, int fa, int p) {
  a[c[u]]++;
  if (a[c[u]] > mx)
    mx = a[c[u]], sum = c[u];
  else if (a[c[u]] == mx)
    sum += c[u];
  for (int i = head[u]; i != -1; i = g[i].nxt) {
    int v = g[i].to;
    if (v == fa || v == p) continue;
    DFS(v, u, p);
  }
}
void dfs(int u, int fa) {
  for (int i = head[u]; i != -1; i = g[i].nxt) {
    int v = g[i].to;
    if (v == fa || v == son[u]) continue;
    dfs(v, u);
    solve(v, u);
    mx = 0, sum = 0;
  }
  if (son[u]) dfs(son[u], u);
  DFS(u, fa, son[u]);
  ans[u] = sum;
}
int main() {
  memset(head, -1, sizeof(head));
  tot = 0;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &c[i]);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    add(u, v);
    add(v, u);
  }
  pre_dfs(1, 0);
  memset(a, 0, sizeof(a));
  mx = 0, sum = 0;
  dfs(1, 0);
  for (int i = 1; i <= n; i++) printf("%lld ", ans[i]);
  return 0;
}
