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 maxn = 100100;
int n, Mx;
int color[maxn], siz[maxn], son[maxn], cnt[maxn];
long long ans[maxn], sum;
vector<int> G[maxn];
void dfs(int u, int fa) {
  siz[u] = 1;
  for (int i = 0; i < G[u].size(); ++i) {
    int v = G[u][i];
    if (v == fa) continue;
    dfs(v, u);
    siz[u] += siz[v];
    if (siz[v] > siz[son[u]]) son[u] = v;
  }
}
void add(int u, int fa, int theson, int val) {
  cnt[color[u]] += val;
  if (cnt[color[u]] > Mx)
    Mx = cnt[color[u]], sum = color[u];
  else if (cnt[color[u]] == Mx)
    sum += color[u];
  for (int i = 0; i < G[u].size(); ++i) {
    int v = G[u][i];
    if (v == fa || v == theson) continue;
    add(v, u, theson, val);
  }
}
void dfs(int u, int fa, int val) {
  for (int i = 0; i < G[u].size(); ++i) {
    int v = G[u][i];
    if (v == fa) continue;
    if (v != son[u]) dfs(v, u, 0);
  }
  if (son[u]) dfs(son[u], u, 1);
  add(u, fa, son[u], 1);
  ans[u] = sum;
  if (val == 0) add(u, fa, 0, -1), sum = 0, Mx = 0;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &color[i]);
  for (int i = 1; i < n; ++i) {
    int u, v;
    scanf("%d%d", &u, &v);
    G[u].push_back(v);
    G[v].push_back(u);
  }
  dfs(1, 0);
  dfs(1, 0, 0);
  for (int i = 1; i <= n; ++i) printf("%lld ", ans[i]);
}
