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 long long N = 1e5 + 10;
inline long long read() {
  long long x = 0, f = 1;
  char ch = getchar();
  while (!isdigit(ch)) {
    if (ch == '-') f = -f;
    ch = getchar();
  }
  while (isdigit(ch)) {
    x = x * 10 + ch - 48;
    ch = getchar();
  }
  return x * f;
}
long long now[N], pre[N << 1], son[N << 1], tot;
void put(long long x, long long y) {
  pre[++tot] = now[x];
  now[x] = tot;
  son[tot] = y;
}
long long n;
long long c[N], child[N], sz[N], fa[N];
long long ans[N], cnt[N], sum, num, pson;
void dfs1(long long x, long long f) {
  sz[x] = 1;
  fa[x] = f;
  long long maxson = 0;
  for (long long i = now[x]; i; i = pre[i]) {
    long long v = son[i];
    if (v == f) continue;
    dfs1(v, x);
    sz[x] += sz[v];
    if (sz[v] > maxson) child[x] = v, maxson = sz[v];
  }
}
void change(long long x, long long val) {
  cnt[c[x]] += val;
  if (cnt[c[x]] > num)
    num = cnt[c[x]], sum = c[x];
  else if (cnt[c[x]] == num)
    sum += c[x];
  for (long long i = now[x]; i; i = pre[i]) {
    long long v = son[i];
    if (v == fa[x] || v == pson) continue;
    change(v, val);
  }
}
void dfs2(long long x, long long opt) {
  for (long long i = now[x]; i; i = pre[i]) {
    long long v = son[i];
    if (v == fa[x] || v == child[x]) continue;
    dfs2(v, 0);
  }
  if (child[x]) dfs2(child[x], 1), pson = child[x];
  change(x, 1);
  ans[x] = sum;
  pson = 0;
  if (!opt) change(x, -1), sum = num = 0;
}
signed main() {
  n = read();
  long long rt = 1;
  for (long long i = 1; i <= n; i++) c[i] = read();
  for (long long i = 1, x, y; i < n; i++) {
    x = read(), y = read();
    put(x, y);
    put(y, x);
  }
  dfs1(rt, 0);
  dfs2(rt, 0);
  for (long long i = 1; i <= n; i++) printf("%lld ", ans[i]);
  return 0;
}
