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;
int n, col[100005], c[100005], dfn[100005], rd[100005], ncnt, blo;
long long ans[100005], cnt[100005], mx, cnt1[100005], answ;
vector<int> G[100005];
struct node {
  int l, r, i;
} q[100005];
bool cmp(node x, node y) {
  if (x.l / blo != y.l / blo) return x.l / blo < y.l / blo;
  return x.r < y.r;
}
void dfs(int u, int fa) {
  dfn[u] = rd[u] = ++ncnt;
  c[ncnt] = col[u];
  for (int i = 0; i < G[u].size(); i++) {
    int v = G[u][i];
    if (v != fa) {
      dfs(v, u);
      rd[u] = max(rd[u], rd[v]);
    }
  }
}
void Delete(int x) {
  cnt1[cnt[c[x]]] -= 1ll * c[x];
  cnt[c[x]]--;
  cnt1[cnt[c[x]]] += 1ll * c[x];
  if (cnt[c[x]] + 1 == mx) {
    if (cnt1[mx])
      answ -= c[x];
    else
      answ = cnt1[--mx];
  }
}
void Add(int x) {
  cnt1[cnt[c[x]]] -= 1ll * c[x];
  cnt[c[x]]++;
  cnt1[cnt[c[x]]] += 1ll * c[x];
  if (cnt[c[x]] > mx)
    mx = cnt[c[x]], answ = c[x];
  else if (cnt[c[x]] == mx)
    answ += 1ll * c[x];
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &col[i]);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    G[u].push_back(v);
    G[v].push_back(u);
  }
  dfs(1, 0);
  for (int i = 1; i <= n; i++) q[i].l = dfn[i], q[i].r = rd[i], q[i].i = i;
  blo = sqrt(n);
  sort(q + 1, q + n + 1, cmp);
  int l = 1, r = 0;
  for (int i = 1; i <= n; i++) {
    while (l < q[i].l) Delete(l++);
    while (l > q[i].l) Add(--l);
    while (r < q[i].r) Add(++r);
    while (r > q[i].r) Delete(r--);
    ans[q[i].i] = answ;
  }
  for (int i = 1; i <= n; i++) printf("%lld ", ans[i]);
}
