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 = 1e5 + 5;
int n, col[N], maxx, cnt[N], p[N], eid, Son, mx[N], sz[N];
long long sum, ans[N];
struct edge {
  int v, nxt;
} e[N << 1];
inline void addedge(int u, int v) { e[++eid] = (edge){v, p[u]}, p[u] = eid; }
inline void dfs0(int u, int fa) {
  sz[u] = 1;
  for (int i = p[u]; i; i = e[i].nxt)
    if (e[i].v != fa) {
      int v = e[i].v;
      dfs0(v, u), sz[u] += sz[v];
      if (sz[v] > sz[mx[u]]) mx[u] = v;
    }
}
inline void add(int u, int fa, int d) {
  cnt[col[u]] += d;
  if (cnt[col[u]] > maxx)
    maxx = cnt[col[u]], sum = col[u];
  else if (cnt[col[u]] == maxx)
    sum += col[u];
  for (int i = p[u]; i; i = e[i].nxt)
    if (e[i].v != fa && e[i].v != Son) add(e[i].v, u, d);
}
inline void dfs(int u, int fa, int d) {
  for (int i = p[u]; i; i = e[i].nxt)
    if (e[i].v != fa && e[i].v != mx[u]) dfs(e[i].v, u, 0);
  if (mx[u]) dfs(mx[u], u, 1), Son = mx[u];
  add(u, fa, 1), Son = 0, ans[u] = sum;
  if (!d) add(u, fa, -1), sum = maxx = 0;
}
int main() {
  scanf("%d", &n);
  for (int i = (1); i <= (n); i++) scanf("%d", &col[i]);
  for (int i = (1); i <= (n - 1); i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    addedge(u, v), addedge(v, u);
  }
  dfs0(1, 0), dfs(1, 0, 0);
  for (int i = (1); i <= (n); i++) printf("%lld ", ans[i]);
  return 0;
}
