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 = 200001;
struct node {
  int nxt, to;
} e[N << 1];
int head[N], cnt = 0, siz[N], son[N], fa[N], max_son, col[N], a[N], maxx, n, x,
             y, Son;
long long sum, ans[N];
void add(int u, int v) {
  e[++cnt].nxt = head[u];
  head[u] = cnt;
  e[cnt].to = v;
}
void dfs(int u, int f, int d) {
  fa[u] = f, siz[u] = 1;
  int max_son = 0;
  for (int i = head[u]; i; i = e[i].nxt) {
    if (e[i].to == f) continue;
    dfs(e[i].to, u, d + 1);
    siz[u] += siz[e[i].to];
    if (siz[e[i].to] > max_son) max_son = siz[e[i].to], son[u] = e[i].to;
  }
}
void calc(int u, int val) {
  col[a[u]] += val;
  if (col[a[u]] > maxx)
    maxx = col[a[u]], sum = a[u];
  else if (col[a[u]] == maxx)
    sum += (long long)a[u];
  for (int i = head[u]; i; i = e[i].nxt) {
    if (e[i].to == fa[u] || e[i].to == Son) continue;
    calc(e[i].to, val);
  }
}
void dfs2(int u, int opt) {
  for (int i = head[u]; i; i = e[i].nxt) {
    if (e[i].to == son[u] || e[i].to == fa[u]) continue;
    dfs2(e[i].to, 0);
  }
  if (son[u]) dfs2(son[u], 1), Son = son[u];
  calc(u, 1);
  ans[u] = sum;
  Son = 0;
  if (!opt) calc(u, -1), sum = maxx = 0;
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i < n; i++) {
    scanf("%d%d", &x, &y);
    add(x, y), add(y, x);
  }
  dfs(1, 0, 1);
  dfs2(1, 0);
  for (int i = 1; i <= n; i++) printf("%lld ", ans[i]);
  return 0;
}
