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;
inline char gc() { return getchar(); }
inline int read() {
  int ans = 0, fh = 1;
  char ch = gc();
  while (ch < '0' || ch > '9') {
    if (ch == '-') fh = -1;
    ch = gc();
  }
  while (ch >= '0' && ch <= '9')
    ans = (ans << 1) + (ans << 3) + ch - '0', ch = gc();
  return ans * fh;
}
const int maxn = 1e5 + 100, maxm = maxn << 1;
int n, head[maxn], nex[maxm], v[maxm], num = 1;
int son[maxn], siz[maxn];
int c[maxn], maxc, cnt[maxn], big[maxn];
long long ans[maxn];
pair<int, long long> f[maxn];
void add(int x, int y) {
  v[++num] = y;
  nex[num] = head[x];
  head[x] = num;
  v[++num] = x;
  nex[num] = head[y];
  head[y] = num;
}
void dfs1(int x, int fa) {
  siz[x] = 1;
  f[0].first++, f[0].second += c[x];
  for (int i = head[x]; i; i = nex[i]) {
    int y = v[i];
    if (y == fa) continue;
    dfs1(y, x);
    if (siz[son[x]] < siz[y]) son[x] = y;
    siz[x] += siz[y];
  }
}
void updata(int x, int fa, int ms) {
  int &cc = cnt[c[x]];
  f[cc].first--, f[cc].second -= c[x];
  cc += ms;
  f[cc].first++, f[cc].second += c[x];
  if (ms > 0)
    maxc = max(maxc, cc);
  else if (!f[maxc].first)
    maxc--;
  for (int i = head[x]; i; i = nex[i])
    if (v[i] != fa && !big[v[i]]) updata(v[i], x, ms);
}
void dsu(int x, int fa, int ms) {
  for (int i = head[x]; i; i = nex[i])
    if (v[i] != fa && v[i] != son[x]) dsu(v[i], x, 0);
  if (x == 5) x = 5;
  if (son[x]) dsu(son[x], x, 1), big[son[x]] = 1;
  updata(x, fa, 1);
  ans[x] = f[maxc].second;
  big[son[x]] = 0;
  if (!ms) updata(x, fa, -1);
}
int main() {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &c[i]);
  int x, y;
  for (int i = 1; i < n; i++) scanf("%d%d", &x, &y), add(x, y);
  dfs1(1, 1);
  dsu(1, 1, 1);
  for (int i = 1; i <= n; i++) printf("%I64d ", ans[i]);
  printf("\n");
  return 0;
}
