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 int N = 2e5 + 105;
const int mod = 1e9 + 7;
const double Pi = acos(-1.0);
const long long INF = 1e18;
const int G = 3, Gi = 332748118;
long long qpow(long long a, long long b) {
  long long res = 1;
  while (b) {
    if (b & 1) res = (res * a) % mod;
    a = (a * a) % mod;
    b >>= 1;
  }
  return res;
}
long long gcd(long long a, long long b) { return b ? gcd(b, a % b) : a; }
int n, m, A, B;
int head[N], cnt = 0;
int to[N << 1], nxt[N << 1];
int son[N], siz[N], dfn[N];
int col[N];
long long ans[N], sum;
int num[N], Max = 0, tson;
void add(int u, int v) {
  to[cnt] = v, nxt[cnt] = head[u], head[u] = cnt++;
  to[cnt] = u, nxt[cnt] = head[v], head[v] = cnt++;
}
void dfs1(int u, int pre) {
  siz[u] = 1;
  int maxx = -1;
  for (int i = head[u]; i != -1; i = nxt[i]) {
    int v = to[i];
    if (v == pre) continue;
    dfs1(v, u);
    siz[u] += siz[v];
    if (siz[v] > maxx) {
      maxx = siz[v];
      son[u] = v;
    }
  }
}
void Addval(int u, int pre, int val) {
  num[col[u]] += val;
  if (num[col[u]] > Max)
    Max = num[col[u]], sum = col[u];
  else if (num[col[u]] == Max)
    sum += col[u];
  for (int i = head[u]; i != -1; i = nxt[i]) {
    int v = to[i];
    if (v == pre || v == tson) continue;
    Addval(v, u, val);
  }
}
void dfs(int u, int pre, int opt) {
  for (int i = head[u]; i != -1; i = nxt[i]) {
    int v = to[i];
    if (v == pre) continue;
    if (v != son[u]) dfs(v, u, 0);
  }
  if (son[u]) dfs(son[u], u, 1), tson = son[u];
  Addval(u, pre, 1), tson = 0;
  ans[u] = sum;
  if (!opt) {
    Addval(u, pre, -1);
    sum = 0, Max = 0;
  }
}
int main() {
  scanf("%d", &n);
  cnt = 0;
  for (int i = 0; i <= n; ++i) head[i] = -1;
  for (int i = 1; i <= n; ++i) scanf("%d", &col[i]);
  for (int i = 1; i < n; ++i) {
    int x, y;
    scanf("%d%d", &x, &y);
    add(x, y);
  }
  dfs1(1, 0);
  dfs(1, 0, 0);
  for (int i = 1; i <= n; ++i) {
    printf("%lld", ans[i]);
    if (i == n)
      printf("\n");
    else
      printf(" ");
  }
  return 0;
}
