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 n, m;
long long a[100010] = {};
struct rode {
  long long y, next;
} e[100010 << 1] = {};
long long fl[100010] = {}, len = 0;
long long si[100010] = {}, d[100010] = {};
long long hs[100010] = {};
inline void inc(long long x, long long y) {
  e[++len].y = y;
  e[len].next = fl[x];
  fl[x] = len;
}
void dfs(long long u, long long fa) {
  si[u] = 1;
  long long maxn = 0;
  for (long long i = fl[u]; i; i = e[i].next)
    if (e[i].y != fa) {
      long long v = e[i].y;
      d[v] = d[u] + 1;
      dfs(v, u);
      si[u] += si[v];
      if (si[v] > maxn) maxn = si[v], hs[u] = v;
    }
}
long long ans[100010] = {};
long long cnt[100010] = {};
long long sum = 0, maxx = 0;
void merge(long long u, long long fa, long long val, long long v) {
  cnt[a[u]] += val;
  if (cnt[a[u]] > maxx)
    maxx = cnt[a[u]], sum = a[u];
  else if (cnt[a[u]] == maxx)
    sum += a[u];
  for (long long i = fl[u]; i; i = e[i].next)
    if (e[i].y != fa && e[i].y != v) {
      long long vv = e[i].y;
      merge(vv, u, val, v);
    }
}
void dsu(long long u, long long fa, long long op) {
  for (long long i = fl[u]; i; i = e[i].next)
    if (e[i].y != fa && e[i].y != hs[u]) {
      long long v = e[i].y;
      dsu(v, u, 0);
    }
  if (hs[u]) dsu(hs[u], u, 1);
  merge(u, fa, 1, hs[u]);
  ans[u] = sum;
  if (op == 0) merge(u, fa, -1, 1), sum = maxx = 0;
}
signed main() {
  scanf("%lld", &n);
  for (long long i = 1; i <= n; ++i) scanf("%lld", &a[i]);
  for (long long i = 1; i < n; ++i) {
    long long x, y;
    scanf("%lld %lld", &x, &y);
    inc(x, y);
    inc(y, x);
  }
  dfs(1, 0);
  dsu(1, 0, 0);
  for (long long i = 1; i <= n; ++i) printf("%lld ", ans[i]);
  return 0;
}
