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 = 100000 + 7;
int n;
int val[N];
struct edge {
  int to, nex;
} e[N << 1];
int fir[N], eid;
int siz[N], fa[N], son[N];
int dfn[N], pos[N], inx;
int cnt[N], res[N];
long long mul[N], ans[N];
void addedge(int u, int v) {
  e[++eid] = (edge){v, fir[u]};
  fir[u] = eid;
}
void hld(int s) {
  siz[s] = 1;
  pos[dfn[s] = ++inx] = s;
  for (int i = fir[s]; i; i = e[i].nex)
    if (e[i].to != fa[s])
      fa[e[i].to] = s, hld(e[i].to), siz[s] += siz[e[i].to],
      son[s] = siz[e[i].to] > siz[son[s]] ? e[i].to : son[s];
}
void dfs(int s, int rem) {
  for (int i = fir[s]; i; i = e[i].nex)
    if (e[i].to != fa[s] && e[i].to != son[s]) dfs(e[i].to, 0);
  if (son[s]) dfs(son[s], 1);
  int c = val[s];
  mul[cnt[c]] -= c, mul[++cnt[c]] += c, res[s] = max(res[son[s]], cnt[c]);
  for (int i = fir[s]; i; i = e[i].nex) {
    if (e[i].to == fa[s] || e[i].to == son[s]) continue;
    for (int j = dfn[e[i].to], c; j < dfn[e[i].to] + siz[e[i].to]; ++j)
      c = val[pos[j]], mul[cnt[c]] -= c, mul[++cnt[c]] += c,
      res[s] = max(res[s], cnt[c]);
  }
  ans[s] = mul[res[s]];
  if (rem) return;
  memset(cnt, 0, sizeof(int) * (n + 2));
  memset(mul, 0, sizeof(long long) * (res[s] + 2));
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", val + i);
  for (int i = 1, u, v; i < n; ++i)
    scanf("%d%d", &u, &v), addedge(u, v), addedge(v, u);
  hld(1);
  dfs(1, 1);
  for (int i = 1; i <= n; ++i) printf("%I64d ", ans[i]);
  return 0;
}
