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 + 10;
vector<int> g[N];
int n, c[N], sz[N], num[N], ma[N];
long long sum[N], ans[N];
void getsz(int u, int fa) {
  int siz = g[u].size();
  sz[u] = 1;
  for (int i = 0; i < siz; i++) {
    int v = g[u][i];
    if (v == fa) continue;
    getsz(v, u);
    sz[u] += sz[v];
  }
}
bool big[N];
void dfs1(int u, int fa, int x) {
  sum[num[c[u]]] -= c[u];
  num[c[u]] += x;
  ma[u] = num[c[u]];
  sum[num[c[u]]] += c[u];
  int siz = g[u].size();
  for (int i = 0; i < siz; i++) {
    int v = g[u][i];
    if (v == fa) continue;
    if (big[v]) {
      ma[u] = max(ma[u], ma[v]);
      continue;
    }
    dfs1(v, u, x);
    ma[u] = max(ma[v], ma[u]);
  }
}
void dfs(int u, int fa, bool keep) {
  int mx = -1, bigc = -1;
  int siz = g[u].size();
  for (int i = 0; i < siz; i++) {
    int v = g[u][i];
    if (v == fa) continue;
    if (sz[v] > mx) bigc = v, mx = sz[v];
  }
  for (int i = 0; i < siz; i++) {
    int v = g[u][i];
    if (v == fa || v == bigc) continue;
    dfs(v, u, 0);
  }
  if (bigc != -1) {
    dfs(bigc, u, 1);
    big[bigc] = 1;
  }
  dfs1(u, fa, 1);
  ans[u] = sum[ma[u]];
  if (bigc != -1) big[bigc] = 0;
  if (keep == 0) dfs1(u, fa, -1);
}
int main(void) {
  scanf("%d", &n);
  for (int i = 1; i <= n; i++) scanf("%d", &c[i]);
  for (int i = 1; i < n; i++) {
    int u, v;
    scanf("%d%d", &u, &v);
    g[u].push_back(v);
    g[v].push_back(u);
  }
  getsz(1, 0);
  dfs(1, 0, 0);
  for (int i = 1; i <= n; i++) printf("%lld%c", ans[i], i == n ? '\n' : ' ');
  return 0;
}
