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 = 1e5 + 5;
long long res = 0, mx = 0;
long long hd[N], cnt = 0, T = 0;
struct Eg {
  long long v, nxt;
} eg[N * 2];
long long sz[N], son[N];
long long c[N], s[N], ans[N];
void add(long long u, long long v) { eg[++cnt] = (Eg){v, hd[u]}, hd[u] = cnt; }
void dfs1(long long x, long long fa) {
  sz[x] = 1;
  for (int i = hd[x]; i; i = eg[i].nxt)
    if (eg[i].v != fa) {
      dfs1(eg[i].v, x);
      sz[x] += sz[eg[i].v];
      if (sz[son[x]] < sz[eg[i].v]) son[x] = eg[i].v;
    }
}
void calc(long long x, long long fa, long long Son) {
  s[c[x]]++;
  if (s[c[x]] > mx)
    mx = s[c[x]], res = c[x];
  else if (s[c[x]] == mx)
    res += c[x];
  for (int i = hd[x]; i; i = eg[i].nxt)
    if (eg[i].v != fa && eg[i].v != Son) calc(eg[i].v, x, Son);
}
void init(long long x, long long fa) {
  s[c[x]]--;
  for (int i = hd[x]; i; i = eg[i].nxt)
    if (eg[i].v != fa) init(eg[i].v, x);
}
void dfs2(long long x, long long fa) {
  for (int i = hd[x]; i; i = eg[i].nxt)
    if (eg[i].v != fa && eg[i].v != son[x])
      dfs2(eg[i].v, x), res = mx = 0, init(eg[i].v, x);
  if (son[x]) dfs2(son[x], x);
  calc(x, fa, son[x]), ans[x] = res;
}
int main() {
  long long n;
  scanf("%I64d", &n);
  for (int i = (1); i <= (n); ++i) scanf("%I64d", &c[i]);
  for (int i = (1); i <= (n - 1); ++i) {
    long long u, v;
    scanf("%I64d%I64d", &u, &v);
    add(u, v), add(v, u);
  }
  dfs1(1, 0), dfs2(1, 0);
  for (int i = (1); i <= (n); ++i) printf("%I64d ", ans[i]);
  return 0;
}
