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;
char buf[1 << 20], *p1, *p2;
template <typename T>
inline void read(T &t) {
  register int v =
      (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 20, stdin), p1 == p2)
           ? EOF
           : *p1++);
  T f = 1;
  t = 0;
  while (!isdigit(v)) {
    if (v == '-') f = -1;
    v = (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 20, stdin), p1 == p2)
             ? EOF
             : *p1++);
  }
  while (isdigit(v)) {
    t = t * 10 + v - 48;
    v = (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 20, stdin), p1 == p2)
             ? EOF
             : *p1++);
  }
  t *= f;
}
template <typename T, typename... Args>
inline void read(T &t, Args &...args) {
  read(t);
  read(args...);
}
const int N = 100001;
const int M = 200010;
template <typename T>
inline T min(T x, T y) {
  return x < y ? x : y;
}
template <typename T>
inline T max(T x, T y) {
  return x > y ? x : y;
}
int to[M], hd[N], nxt[M], tot;
inline void add(int u, int v) {
  to[++tot] = v;
  nxt[tot] = hd[u];
  hd[u] = tot;
}
inline void addedge(int u, int v) { add(u, v), add(v, u); }
long long son[N], siz[N], cnt[N], col[N], sum, Son, maxx, n, ans[N];
void dfs1(int u, int f) {
  siz[u] = 1;
  son[u] = 0;
  for (int i = hd[u]; i; i = nxt[i]) {
    int v = to[i];
    if (v != f) {
      dfs1(v, u);
      siz[u] += siz[v];
      if (siz[v] > siz[son[u]]) {
        son[u] = v;
      }
    }
  }
}
void add(int u, int f, int val) {
  cnt[col[u]] += val;
  if (cnt[col[u]] > maxx)
    maxx = cnt[col[u]], sum = col[u];
  else if (cnt[col[u]] == maxx)
    sum += col[u];
  for (int i = hd[u]; i; i = nxt[i]) {
    int v = to[i];
    if (v != f && v != Son) {
      add(v, u, val);
    }
  }
}
void dfs2(int u, int f, int opt) {
  for (int i = hd[u]; i; i = nxt[i]) {
    int v = to[i];
    if (v != f && v != son[u]) {
      dfs2(v, u, 0);
    }
  }
  if (son[u]) dfs2(son[u], u, 1), Son = son[u];
  add(u, f, 1), Son = 0;
  ans[u] = sum;
  if (opt == 0) add(u, f, -1), sum = maxx = 0;
}
signed main() {
  read(n);
  for (int i = 1; i <= n; ++i) read(col[i]);
  for (int i = 1; i < n; ++i) {
    int x, y;
    read(x, y);
    addedge(x, y);
  }
  dfs1(1, 0);
  dfs2(1, 0, 0);
  for (int i = 1; i <= n; ++i) printf("%lld ", ans[i]);
  return 0;
}
