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 M = 1e5, MAXN = 1e5 + 5;
long long V[MAXN * 40], ans[MAXN];
int Mx[MAXN * 40], rt[MAXN], tot = 0, c[MAXN], n, ls[MAXN * 40], rs[MAXN * 40];
vector<int> G[MAXN];
void push_up(int o) {
  if (Mx[ls[o]] == Mx[rs[o]]) {
    Mx[o] = Mx[ls[o]];
    V[o] = V[ls[o]] + V[rs[o]];
  } else if (Mx[ls[o]] > Mx[rs[o]]) {
    Mx[o] = Mx[ls[o]];
    V[o] = V[ls[o]];
  } else {
    Mx[o] = Mx[rs[o]];
    V[o] = V[rs[o]];
  }
}
void ud(int &o, int l, int r, int p) {
  int mid = (l + r) / 2;
  if (!o) o = ++tot;
  if (l == r) {
    V[o] = l;
    Mx[o]++;
    return;
  }
  if (p <= mid)
    ud(ls[o], l, mid, p);
  else
    ud(rs[o], mid + 1, r, p);
  push_up(o);
}
int merge(int a, int b, int l, int r) {
  int mid = (l + r) / 2;
  if (!a || !b) return a | b;
  int z = ++tot;
  ls[z] = merge(ls[a], ls[b], l, mid);
  rs[z] = merge(rs[a], rs[b], mid + 1, r);
  if (ls[z] || rs[z])
    push_up(z);
  else {
    Mx[z] = Mx[a] + Mx[b];
    V[z] = l;
  }
  return z;
}
void dfs(int u, int fa) {
  ud(rt[u], 1, M, c[u]);
  for (auto v : G[u]) {
    if (v == fa) continue;
    dfs(v, u);
    rt[u] = merge(rt[u], rt[v], 1, M);
  }
  ans[u] = V[rt[u]];
}
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) {
    scanf("%d", &c[i]);
  }
  for (int i = 1; i <= n - 1; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    G[u].push_back(v);
    G[v].push_back(u);
  }
  dfs(1, 0);
  for (int i = 1; i <= n; i++) printf("%lld ", ans[i]);
}
