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;
long long read() {
  char c = getchar();
  bool flag = 0;
  long long x = 0;
  while (c < '0' || c > '9') {
    if (c == '-') flag = 1;
    c = getchar();
  }
  while (c >= '0' && c <= '9') {
    x = (x << 3) + (x << 1) + c - '0';
    c = getchar();
  }
  return flag ? -x : x;
}
const long long N = 1e5 + 10;
struct edge {
  long long to, nxt;
} ln[N * 2];
long long idx[N], topln;
void add_edge(long long u, long long v) {
  ln[++topln] = (edge){v, idx[u]}, idx[u] = topln;
  ln[++topln] = (edge){u, idx[v]}, idx[v] = topln;
}
long long siz[N], son[N], dfn[N], dfn_d[N], js = 0;
void dfs1(long long x, long long fa) {
  dfn[x] = ++js, dfn_d[js] = x, siz[x] = 1;
  long long mx = 0;
  for (long long i = idx[x]; i; i = ln[i].nxt) {
    if (ln[i].to == fa) continue;
    dfs1(ln[i].to, x);
    siz[x] += siz[ln[i].to];
    if (siz[ln[i].to] > mx) son[x] = ln[i].to, mx = siz[ln[i].to];
  }
}
long long pt[N], c[N], cnt[N], sum[N], mx = 0;
void Add(long long x) {
  sum[cnt[c[x]]] -= c[x];
  cnt[c[x]]++;
  sum[cnt[c[x]]] += c[x];
  mx = max(cnt[c[x]], mx);
}
void Del(long long x) {
  sum[cnt[c[x]]] -= c[x];
  cnt[c[x]]--;
  sum[cnt[c[x]]] += c[x];
  if (mx == cnt[c[x]] + 1 && sum[cnt[c[x]] + 1] == 0) mx--;
}
void dfs2(long long x, long long fa, long long sve) {
  for (long long i = idx[x]; i; i = ln[i].nxt) {
    if (ln[i].to == fa || ln[i].to == son[x]) continue;
    dfs2(ln[i].to, x, 0);
  }
  if (son[x]) dfs2(son[x], x, 1);
  for (long long i = idx[x]; i; i = ln[i].nxt) {
    if (ln[i].to == fa || ln[i].to == son[x]) continue;
    for (long long j = dfn[ln[i].to]; j <= dfn[ln[i].to] + siz[ln[i].to] - 1;
         j++) {
      Add(dfn_d[j]);
    }
  }
  Add(x);
  pt[x] = sum[mx];
  if (sve) return;
  for (long long i = dfn[x]; i <= dfn[x] + siz[x] - 1; i++) {
    Del(dfn_d[i]);
  }
}
long long vis[N];
int main() {
  long long n = read();
  for (long long i = 1; i <= n; i++) c[i] = read();
  for (long long i = 1; i < n; i++) {
    long long u = read(), v = read();
    add_edge(u, v);
  }
  dfs1(1, 0);
  for (long long i = 1; i <= n; i++) {
    if (vis[c[i]]) continue;
    vis[c[i]] = 1, sum[0] += c[i];
  }
  dfs2(1, 0, 1);
  for (long long i = 1; i <= n; i++) printf("%lld ", pt[i]);
}
