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;
struct node {
  int l, r, max;
  long long ans;
} tr[2000001];
struct node1 {
  int x, y, ne;
} a[200001];
int len = 0, la[100001];
long long w[100001], t[100001];
long long root[100001];
int n;
void read(int &num) {
  char ch;
  while (ch = getchar(), ch < '!')
    ;
  num = ch - 48;
  while (ch = getchar(), ch > '!') num = num * 10 + ch - 48;
}
void add(int x, int y) {
  len++;
  int k = len;
  a[k].x = x, a[k].y = y;
  a[k].ne = la[x], la[x] = len;
}
int find(int k1, int k2, int l, int r) {
  if ((k1 == 0) || (k2 == 0)) return k1 + k2;
  if (l == r) {
    tr[k1].max = tr[k1].max + tr[k2].max;
    return k1;
  }
  int mid = (l + r) >> 1;
  tr[k1].l = find(tr[k1].l, tr[k2].l, l, mid);
  tr[k1].r = find(tr[k1].r, tr[k2].r, mid + 1, r);
  int ls = tr[k1].l, rs = tr[k1].r;
  if (tr[ls].max > tr[rs].max) {
    tr[k1].max = tr[ls].max;
    tr[k1].ans = tr[ls].ans;
  }
  if (tr[ls].max < tr[rs].max) {
    tr[k1].max = tr[rs].max;
    tr[k1].ans = tr[rs].ans;
  }
  if (tr[ls].max == tr[rs].max) {
    tr[k1].max = tr[ls].max;
    tr[k1].ans = tr[ls].ans + tr[rs].ans;
  }
  return k1;
}
void add2(int k, int l, int r, int pos) {
  if (l == r) {
    tr[k].max++;
    tr[k].ans = l;
    return;
  }
  int mid = (l + r) >> 1;
  if (pos <= mid) {
    if (tr[k].l == 0) {
      len++;
      tr[k].l = len;
    }
    add2(tr[k].l, l, mid, pos);
  } else {
    if (tr[k].r == 0) {
      len++;
      tr[k].r = len;
    }
    add2(tr[k].r, mid + 1, r, pos);
  }
  int ls = tr[k].l, rs = tr[k].r;
  if (tr[ls].max > tr[rs].max) {
    tr[k].max = tr[ls].max;
    tr[k].ans = tr[ls].ans;
  }
  if (tr[ls].max < tr[rs].max) {
    tr[k].max = tr[rs].max;
    tr[k].ans = tr[rs].ans;
  }
  if (tr[ls].max == tr[rs].max) {
    tr[k].max = tr[ls].max;
    tr[k].ans = tr[ls].ans + tr[rs].ans;
  }
}
void go(int x, int fa) {
  for (int k = la[x]; k; k = a[k].ne) {
    int y = a[k].y;
    if (y == fa) continue;
    go(y, x);
    root[x] = find(root[x], root[y], 1, n);
  }
  if (root[x] == 0) {
    len++;
    root[x] = len;
  }
  add2(root[x], 1, n, w[x]);
  t[x] = tr[root[x]].ans;
}
int main() {
  int x, y;
  read(n);
  memset(root, 0, sizeof(root));
  for (int i = 1; i <= n; i++) scanf("%lld", &w[i]);
  for (int i = 1; i <= n - 1; i++) {
    read(x), read(y);
    add(x, y), add(y, x);
  }
  go(1, 0);
  for (int i = 1; i <= n; i++) printf("%lld ", t[i]);
  return 0;
}
