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 long long MAX_N = 2e6 + 5;
const long long MOD = 1e9 + 7;
long long N, M, K;
long long head[MAX_N];
long long tot = 0;
struct Edge {
  long long to, nxt;
} edge[MAX_N];
void addEdge(long long u, long long v) {
  edge[tot].nxt = head[u];
  edge[tot].to = v;
  head[u] = tot++;
  edge[tot].nxt = head[v];
  edge[tot].to = u;
  head[v] = tot++;
}
long long arr[MAX_N];
long long son[MAX_N];
long long sz[MAX_N];
void dfs1(long long u, long long from) {
  sz[u] = 1;
  long long v;
  for (long long i = head[u]; ~i; i = edge[i].nxt) {
    if ((v = edge[i].to) == from) continue;
    dfs1(v, u);
    sz[u] += sz[v];
    if (sz[v] > sz[son[u]]) {
      son[u] = v;
    }
  }
}
long long col[MAX_N];
long long maxx = 0;
long long cnt = 0;
long long ans[MAX_N];
void calc(long long u, long long from) {
  if (col[arr[u]] + 1 > maxx) {
    cnt = arr[u];
  } else if (col[arr[u]] + 1 == maxx) {
    cnt += arr[u];
  }
  maxx = max(maxx, ++col[arr[u]]);
  for (long long i = head[u]; ~i; i = edge[i].nxt) {
    if (edge[i].to == from) continue;
    calc(edge[i].to, u);
  }
}
void del(long long u, long long from) {
  --col[arr[u]];
  for (long long i = head[u]; ~i; i = edge[i].nxt) {
    if (edge[i].to == from) continue;
    del(edge[i].to, u);
  }
}
void dsu(long long u, long long opt, long long from) {
  long long v;
  for (long long i = head[u]; ~i; i = edge[i].nxt) {
    if ((v = edge[i].to) == from || v == son[u]) continue;
    dsu(v, 1, u);
  }
  if (son[u]) dsu(son[u], 0, u);
  if (col[arr[u]] + 1 > maxx) {
    cnt = arr[u];
  } else if (col[arr[u]] + 1 == maxx) {
    cnt += arr[u];
  }
  maxx = max(maxx, ++col[arr[u]]);
  for (long long i = head[u]; ~i; i = edge[i].nxt) {
    if ((v = edge[i].to) == from || v == son[u]) continue;
    calc(v, u);
  }
  ans[u] = cnt;
  if (opt) {
    del(u, from);
    maxx = 0;
    cnt = 0;
  }
}
void init() {
  memset(head, -1, sizeof head);
  tot = 0;
}
void solve() {
  init();
  scanf("%lld", &N);
  for (long long i = 1; i <= N; ++i) {
    scanf("%lld", &arr[i]);
  }
  long long u, v;
  for (long long i = 1; i < N; ++i) {
    scanf("%lld%lld", &u, &v);
    addEdge(u, v);
  }
  dfs1(1, 0);
  dsu(1, 0, 0);
  for (long long i = 1; i <= N; ++i) {
    printf("%lld ", ans[i]);
  }
}
signed main() {
  long long T = 1;
  while (T--) solve();
  return 0;
}
