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;
int gi() {
  int x = 0, c = getchar();
  bool f = 0;
  for (; !isdigit(c); c = getchar())
    if (c == '-') f = 1;
  for (; isdigit(c); c = getchar()) x = x * 10 + (c & 15);
  return f ? -x : x;
}
const int N = 100005;
int n, c[N], hd[N], cnt;
struct edge {
  int to, nxt;
  edge() {}
  edge(int to_, int nxt_) : to(to_), nxt(nxt_) {}
} e[N << 1];
int fa[N], dep[N], siz[N], son[N];
long long ans[N];
struct _A {
  int cnt[N], stk[N], top, mx;
  long long ans;
  void add(int x) {
    stk[++top] = x;
    cnt[x]++;
    if (cnt[x] > mx) {
      mx = cnt[x];
      ans = x;
    } else if (cnt[x] == mx) {
      ans += x;
    }
  }
  void clear() {
    while (top) {
      cnt[stk[top--]]--;
    }
    mx = 0;
    ans = 0;
  }
  long long query() { return ans; }
} A;
void adde(int u, int v) {
  e[++cnt] = edge(v, hd[u]);
  hd[u] = cnt;
  e[++cnt] = edge(u, hd[v]);
  hd[v] = cnt;
}
void dfs1(int u) {
  siz[u] = 1;
  for (int i = hd[u]; i; i = e[i].nxt) {
    int v = e[i].to;
    if (v == fa[u]) continue;
    fa[v] = u;
    dep[v] = dep[u] + 1;
    dfs1(v);
    siz[u] += siz[v];
    if (siz[v] > siz[son[u]]) {
      son[u] = v;
    }
  }
}
void dfs3(int u) {
  for (int i = hd[u]; i; i = e[i].nxt) {
    int v = e[i].to;
    if (v == fa[u]) continue;
    dfs3(v);
  }
  A.add(c[u]);
}
void dfs2(int u) {
  for (int i = hd[u]; i; i = e[i].nxt) {
    int v = e[i].to;
    if (v == fa[u] || v == son[u]) continue;
    dfs2(v);
    A.clear();
  }
  if (son[u]) dfs2(son[u]);
  for (int i = hd[u]; i; i = e[i].nxt) {
    int v = e[i].to;
    if (v == fa[u] || v == son[u]) continue;
    dfs3(v);
  }
  A.add(c[u]);
  ans[u] = A.query();
}
int main() {
  n = gi();
  for (int i = 1; i <= n; ++i) {
    c[i] = gi();
  }
  for (int i = 1; i < n; ++i) {
    adde(gi(), gi());
  }
  dfs1(1);
  dfs2(1);
  for (int i = 1; i <= n; ++i) {
    printf("%lld ", ans[i]);
  }
  return 0;
}
