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 = 100010;
long long n, m, tot, head[N], c[N], son[N], siz[N], cnt[N];
long long sum, ans[N], maxi, _;
struct Edge {
  long long ver, suiv;
} e[N << 1];
inline void lnk(long long x, long long y) {
  e[++tot].ver = y;
  e[tot].suiv = head[x];
  head[x] = tot;
}
inline void dfs1(long long x, long long fa, long long maxson = -1) {
  siz[x] = 1;
  for (long long i = head[x]; i; i = e[i].suiv) {
    long long y = e[i].ver;
    if (y == fa) continue;
    dfs1(y, x);
    siz[x] += siz[y];
    if (siz[y] > maxson) maxson = siz[y], son[x] = y;
  }
}
inline void update(long long x, long long fa, long long gar, long long val) {
  cnt[c[x]] += val;
  if (cnt[c[x]] > maxi)
    maxi = cnt[c[x]], sum = c[x];
  else if (cnt[c[x]] == maxi)
    sum += c[x];
  for (long long i = head[x]; i; i = e[i].suiv) {
    long long y = e[i].ver;
    if (y == fa || y == gar) continue;
    update(y, x, gar, val);
  }
}
inline void dfs2(long long x, long long fa, long long opt) {
  for (long long i = head[x]; i; i = e[i].suiv) {
    long long y = e[i].ver;
    if (y == son[x] || y == fa) continue;
    dfs2(y, x, 0);
  }
  if (son[x]) dfs2(son[x], x, 1), _ = son[x];
  update(x, fa, _, 1);
  _ = 0, ans[x] = sum;
  if (!opt) update(x, fa, _, -1), sum = maxi = 0;
}
signed main() {
  scanf("%lld", &n);
  for (long long i = 1; i <= n; i++) scanf("%lld", &c[i]);
  for (long long i = 1, x, y; i < n; i++) {
    scanf("%lld%lld", &x, &y);
    lnk(x, y), lnk(y, x);
  }
  dfs1(1, 1);
  dfs2(1, 1, 0);
  for (long long i = 1; i <= n; i++) printf("%lld\n", ans[i]);
  return 0;
}
