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 + 10;
int n, c[M];
int num = 0, head[M], End[2 * M], Next[2 * M];
int size[M], son[M], fa[M];
long long ans[M];
struct node {
  int dfn, Max, cnt[M], pre[M];
  long long sum = 0;
  void clear() {
    dfn++;
    Max = sum = 0;
  }
  void inc(int k) {
    if (pre[k] != dfn) cnt[k] = 0, pre[k] = dfn;
    cnt[k]++;
    if (cnt[k] > Max) sum = 0, Max = cnt[k];
    if (cnt[k] == Max) sum += k;
  }
  long long ans() { return sum; }
} A;
inline void add(int x, int y) {
  End[++num] = y;
  Next[num] = head[x];
  head[x] = num;
}
void dfs1(int x) {
  size[x] = 1;
  for (int i = head[x]; i; i = Next[i]) {
    int y = End[i];
    if (y == fa[x]) continue;
    fa[y] = x;
    dfs1(y);
    size[x] += size[y];
    if (size[y] > size[son[x]]) son[x] = y;
  }
}
void calc(int x, node &A) {
  A.inc(c[x]);
  for (int i = head[x]; i; i = Next[i])
    if (End[i] != fa[x]) calc(End[i], A);
}
void dfs(int x, node &A) {
  if (!son[x]) {
    ans[x] = c[x];
    A.inc(c[x]);
    return;
  }
  for (int i = head[x]; i; i = Next[i]) {
    int y = End[i];
    if (y == fa[x] || y == son[x]) continue;
    dfs(y, A);
    A.clear();
  }
  dfs(son[x], A);
  for (int i = head[x]; i; i = Next[i]) {
    int y = End[i];
    if (y == fa[x] || y == son[x]) continue;
    calc(y, A);
  }
  A.inc(c[x]);
  ans[x] = A.ans();
}
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);
    add(x, y);
    add(y, x);
  }
  dfs1(1);
  dfs(1, A);
  for (int i = 1; i <= n; i++) printf("%I64d ", ans[i]);
  return 0;
}
