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 + 10;
const long long mod = 1e9 + 7;
int head[maxn], n, siz[maxn], son[maxn];
int col[maxn], cnt[maxn];
long long ans[maxn], sum, maxc;
int tot = 0, flag;
struct node {
  int next, v;
} ed[maxn * 2];
void add(int u, int v) {
  ed[tot].v = v;
  ed[tot].next = head[u];
  head[u] = tot++;
}
void dfs(int u, int fa) {
  siz[u] = 1;
  for (int i = head[u]; ~i; i = ed[i].next) {
    int v = ed[i].v;
    if (v == fa) continue;
    dfs(v, u);
    siz[u] += siz[v];
    if (siz[v] > siz[son[u]]) son[u] = v;
  }
}
void count(int u, int fa, int val) {
  cnt[col[u]] += val;
  if (cnt[col[u]] > maxc) {
    maxc = cnt[col[u]];
    sum = col[u];
  } else if (cnt[col[u]] == maxc)
    sum += col[u];
  for (int i = head[u]; ~i; i = ed[i].next) {
    int v = ed[i].v;
    if (v == fa || v == flag) continue;
    count(v, u, val);
  }
}
void dfs2(int u, int fa, bool keep) {
  for (int i = head[u]; ~i; i = ed[i].next) {
    int v = ed[i].v;
    if (v == fa || v == son[u]) continue;
    dfs2(v, u, false);
  }
  if (son[u]) {
    dfs2(son[u], u, 1);
    flag = son[u];
  }
  count(u, fa, 1);
  flag = 0;
  ans[u] = sum;
  if (!keep) {
    count(u, fa, -1);
    sum = maxc = 0;
  }
}
int main() {
  cin >> n;
  for (int i = 1; i < n + 1; i++) scanf("%d", &col[i]);
  memset(head, -1, sizeof head);
  for (int i = 0; i < n - 1; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    add(u, v);
    add(v, u);
  }
  dfs(1, 0);
  dfs2(1, 0, 0);
  for (int i = 1; i < n + 1; i++) printf("%lld ", ans[i]);
}
