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 = 2e5 + 10;
int head[N], cent = 0;
int c[N];
struct node {
  int v;
  int next;
} tr[N * 2];
void add(int u, int v) {
  tr[cent].v = v;
  tr[cent].next = head[u];
  head[u] = cent++;
}
int siz[N], son[N];
int Son, cnt[N];
long long ans[N], sum, maxn;
void dfs1(int u, int fa) {
  siz[u] = 1;
  for (int i = head[u]; ~i; i = tr[i].next) {
    int to = tr[i].v;
    if (to == fa) continue;
    dfs1(to, u);
    siz[u] += siz[to];
    if (siz[son[u]] < siz[to]) son[u] = to;
  }
}
void add(int x, int fa, int val) {
  cnt[c[x]] += val;
  if (cnt[c[x]] > maxn) {
    maxn = cnt[c[x]];
    sum = c[x];
  } else if (cnt[c[x]] == maxn)
    sum += c[x];
  for (int i = head[x]; ~i; i = tr[i].next) {
    int to = tr[i].v;
    if (to == fa || to == Son) continue;
    add(to, x, val);
  }
}
void dfs2(int x, int fa, int op) {
  for (int i = head[x]; ~i; i = tr[i].next) {
    int to = tr[i].v;
    if (to == fa || to == son[x]) continue;
    dfs2(to, x, 0);
  }
  if (son[x]) {
    dfs2(son[x], x, 1);
    Son = son[x];
  }
  add(x, fa, 1);
  ans[x] = sum;
  Son = 0;
  if (op == 0) {
    add(x, fa, -1);
    maxn = 0;
    sum = 0;
  }
}
int main() {
  int n;
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &c[i]);
  memset(head, -1, sizeof(head));
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    add(u, v);
    add(v, u);
  }
  dfs1(1, 1);
  dfs2(1, 1, 0);
  for (int i = 1; i < n; i++) printf("%lld ", ans[i]);
  printf("%lld\n", ans[n]);
  return 0;
}
