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;
inline long long read() {
  long long re = 0, k = 1;
  char ch = getchar();
  while (ch > '9' || ch < '0') {
    if (ch == '-') k = -1;
    ch = getchar();
  }
  while (ch <= '9' && ch >= '0') {
    re = re * 10 + ch - 48;
    ch = getchar();
  }
  return re * k;
}
inline void write(long long x) {
  if (x < 0) return putchar('-'), write(-x), void();
  if (x <= 9) return putchar(x + 48), void();
  return write(x / 10), write(x % 10), void();
}
long long sum, head[1000005], tot, col[1000005], t[1000005], ans[1000005], mx,
    n;
struct ss {
  long long node, nxt;
} e[2000005];
void add(long long u, long long v) {
  e[++tot].nxt = head[u];
  e[tot].node = v;
  head[u] = tot;
}
long long son[1000005], sz[1000005];
void dfs(long long x, long long fa) {
  sz[x] = 1;
  long long mxsz = -1;
  for (register long long i = head[x]; i; i = e[i].nxt) {
    long long y = e[i].node;
    if (y == fa) continue;
    dfs(y, x);
    sz[x] += sz[y];
    if (sz[y] > mxsz) son[x] = y, mxsz = sz[y];
  }
}
void getval(long long x, long long fa, long long p) {
  t[col[x]]++;
  if (t[col[x]] > mx)
    mx = t[col[x]], sum = col[x];
  else if (t[col[x]] == mx)
    sum += col[x];
  for (register long long i = head[x]; i; i = e[i].nxt) {
    long long y = e[i].node;
    if (y == fa || y == p) continue;
    getval(y, x, p);
  }
}
void init(long long x, long long fa) {
  t[col[x]]--;
  for (register long long i = head[x]; i; i = e[i].nxt) {
    long long y = e[i].node;
    if (y == fa) continue;
    init(y, x);
  }
}
void dfs2(long long x, long long fa) {
  cerr << x << endl;
  for (register long long i = head[x]; i; i = e[i].nxt) {
    long long y = e[i].node;
    if (y == fa || y == son[x]) continue;
    dfs2(y, x);
    init(y, x);
    sum = mx = 0;
  }
  if (son[x]) {
    dfs2(son[x], x);
  }
  getval(x, fa, son[x]);
  ans[x] = sum;
}
signed main() {
  n = read();
  for (register long long i = 1; i <= n; i++) {
    col[i] = read();
  }
  for (register long long i = 1, u, v; i < n; i++) {
    u = read();
    v = read();
    add(u, v);
    add(v, u);
  }
  dfs(1, 0);
  dfs2(1, 0);
  for (register long long i = 1; i <= n; i++) {
    write(ans[i]);
    putchar(' ');
  }
}
