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 + 5;
struct Edge {
  int to, next;
} e[N << 1];
int head[N], tot;
void add(int u, int v) {
  e[++tot].to = v;
  e[tot].next = head[u];
  head[u] = tot;
}
long long siz[N], son[N];
void dfs(int u, int fath) {
  siz[u] = 1;
  for (int i = head[u]; i; i = e[i].next) {
    int v = e[i].to;
    if (v == fath) continue;
    dfs(v, u);
    siz[u] += siz[v];
    if (siz[v] > siz[son[u]]) son[u] = v;
  }
}
long long col[N], ans[N], cnt[N];
long long big, sum, max_cnt;
void deal(int u, int fath, int val) {
  cnt[col[u]] += val;
  if (cnt[col[u]] > max_cnt && val > 0) {
    max_cnt = cnt[col[u]];
    sum = (long long)col[u];
  } else if (cnt[col[u]] == max_cnt && val > 0) {
    sum += (long long)col[u];
  }
  for (int i = head[u]; i; i = e[i].next) {
    int v = e[i].to;
    if (v != fath && v != big) deal(v, u, val);
  }
}
void dsu(int u, int fath, bool kep) {
  for (int i = head[u]; i; i = e[i].next) {
    int v = e[i].to;
    if (v != fath && v != son[u]) dsu(v, u, 0);
  }
  if (son[u]) {
    dsu(son[u], u, 1);
    big = son[u];
  }
  deal(u, fath, 1);
  big = 0;
  ans[u] = sum;
  if (!kep) {
    deal(u, fath, -1);
    sum = 0;
    max_cnt = 0;
  }
}
int main() {
  long long n;
  scanf("%lld", &n);
  for (int i = 1; i <= n; i++) scanf("%lld", &col[i]);
  for (int i = 1; i < n; i++) {
    long long u, v;
    scanf("%lld%lld", &u, &v);
    add(u, v);
    add(v, u);
  }
  dfs(1, 0);
  dsu(1, 0, 1);
  for (int i = 1; i <= n; i++) printf("%lld%c", ans[i], " \n"[i == n]);
  return 0;
}
