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 = 1e5 + 5;
int n, c[maxn];
vector<int> e[maxn];
int fa[maxn], sz[maxn], son[maxn];
void dfs(int u, int f) {
  fa[u] = f;
  sz[u] = 1;
  for (int i = 0; i < e[u].size(); ++i) {
    int v = e[u][i];
    if (v == f) continue;
    dfs(v, u);
    sz[u] += sz[v];
    if (!son[u] || sz[son[u]] < sz[v]) son[u] = v;
  }
}
int cnt[maxn], top;
long long sum[maxn], ans[maxn];
void Add(int u, int add) {
  sum[cnt[c[u]]] -= 1ll * c[u];
  cnt[c[u]] += add;
  sum[cnt[c[u]]] += 1ll * c[u];
  if (sum[top + 1]) top++;
  if (!sum[top]) top--;
  for (int i = 0; i < e[u].size(); ++i) {
    int v = e[u][i];
    if (v == fa[u]) continue;
    Add(v, add);
  }
}
void solve(int u, int keep) {
  for (int i = 0; i < e[u].size(); ++i) {
    int v = e[u][i];
    if (v == fa[u] || v == son[u]) continue;
    solve(v, 0);
  }
  if (son[u]) solve(son[u], 1);
  for (int i = 0; i < e[u].size(); ++i) {
    int v = e[u][i];
    if (v == fa[u] || v == son[u]) continue;
    Add(v, 1);
  }
  sum[cnt[c[u]]] -= 1ll * c[u];
  cnt[c[u]]++;
  sum[cnt[c[u]]] += 1ll * c[u];
  if (sum[top + 1]) top++;
  if (!sum[top]) top--;
  ans[u] = sum[top];
  if (!keep) Add(u, -1);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &c[i]);
  for (int i = 1; i < n; ++i) {
    int x, y;
    scanf("%d%d", &x, &y);
    e[x].push_back(y);
    e[y].push_back(x);
  }
  dfs(1, 0);
  solve(1, 1);
  for (int i = 1; i <= n; ++i) {
    printf("%lld ", ans[i]);
  }
}
