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 inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double pi = acos(-1.0);
const int dx[] = {-1, 1, 0, 0};
const int dy[] = {0, 0, -1, 1};
int c[100005], size[100005], mxc[100005], cnt[100005];
vector<int> edges[100005], *v[100005];
long long int sum[100005];
void dfs(int cur, int par) {
  size[cur]++;
  for (int nxt : edges[cur]) {
    if (nxt == par) continue;
    dfs(nxt, cur);
    size[cur] += size[nxt];
  }
}
void dfs(int cur, int par, bool keep) {
  int big = 0, mx = 0;
  for (int nxt : edges[cur]) {
    if (nxt == par) continue;
    if (size[nxt] > mx) mx = size[nxt], big = nxt;
  }
  for (int nxt : edges[cur]) {
    if (nxt == par || nxt == big) continue;
    dfs(nxt, cur, 0);
  }
  if (big) {
    dfs(big, cur, 1);
    v[cur] = v[big];
    mxc[cur] = mxc[big];
    sum[cur] = sum[big];
  } else {
    v[cur] = new vector<int>();
  }
  v[cur]->emplace_back(c[cur]);
  cnt[c[cur]]++;
  if (big) {
    if (cnt[c[cur]] > mxc[big]) {
      mxc[cur] = cnt[c[cur]];
      sum[cur] = c[cur];
    } else if (cnt[c[cur]] == mxc[big]) {
      sum[cur] += c[cur];
    }
  } else {
    mxc[cur] = 1;
    sum[cur] = c[cur];
  }
  for (int nxt : edges[cur]) {
    if (nxt == par || nxt == big) continue;
    for (int col : *v[nxt]) {
      v[cur]->emplace_back(col);
      cnt[col]++;
      if (cnt[col] > mxc[cur]) {
        mxc[cur] = cnt[col];
        sum[cur] = col;
      } else if (cnt[col] == mxc[cur]) {
        sum[cur] += col;
      }
    }
  }
  if (!keep)
    for (int col : *v[cur]) cnt[col] = 0;
}
int main() {
  ios_base::sync_with_stdio(0);
  int n, u, w;
  while (cin >> n) {
    for (int i = 1; i <= n; ++i) cin >> c[i];
    for (int i = 1; i < n; ++i) {
      cin >> u >> w;
      edges[u].push_back(w);
      edges[w].push_back(u);
    }
    memset(mxc, 0, sizeof(mxc));
    memset(sum, 0, sizeof(sum));
    memset(size, 0, sizeof(size));
    dfs(1, 0);
    dfs(1, 0, 0);
    for (int i = 1; i <= n; ++i) {
      cout << sum[i] << " ";
      edges[i].clear();
    }
    cout << "\n";
  }
  return 0;
}
