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 maxn = 1e5 + 7;
const int INF = 0x3f3f3f3f;
const int MOD = 1e9 + 7;
struct edge {
  int to, nxt, weight;
} e[maxn * 2];
int n, sz[maxn], sum[maxn];
int dep[maxn], cnt;
int head[maxn], tot;
int parent[maxn], hson[maxn];
void init() {
  tot = 0;
  memset(head, -1, sizeof(head));
}
void add(int u, int v) {
  e[tot].to = v;
  e[tot].nxt = head[u];
  head[u] = tot++;
}
void dfs1(int u, int fa, int root) {
  parent[u] = fa;
  dep[u] = dep[fa] + 1;
  sz[u] = 1;
  for (int i = head[u], v; i != -1; i = e[i].nxt) {
    v = e[i].to;
    if (v == fa) continue;
    dfs1(v, u, root);
    sz[u] += sz[v];
    if (sz[v] > sz[hson[u]]) hson[u] = v;
  }
}
int mx[maxn], c[maxn], colcnt[maxn], Son;
long long ans[maxn];
void cal(int rt, int u, int val) {
  colcnt[c[u]] += val;
  if (val == 1) {
    if (colcnt[c[u]] > mx[rt])
      ans[rt] = c[u], mx[rt] = colcnt[c[u]];
    else if (colcnt[c[u]] == mx[rt])
      ans[rt] += c[u];
  }
  for (int i = head[u], v; i != -1; i = e[i].nxt) {
    v = e[i].to;
    if (v == parent[u] || v == Son) continue;
    cal(rt, v, val);
  }
}
void solve(int u, bool isson) {
  for (int i = head[u], v; i != -1; i = e[i].nxt) {
    v = e[i].to;
    if (v == parent[u] || v == hson[u]) continue;
    solve(v, 0);
  }
  if (hson[u])
    solve(hson[u], 1), Son = hson[u], mx[u] = mx[Son], ans[u] = ans[Son];
  cal(u, u, 1);
  Son = 0;
  if (!isson) cal(u, u, -1);
}
int main() {
  init();
  scanf("%d", &n);
  for (int i = 1; i <= n; ++i) scanf("%d", &c[i]);
  for (int i = 1, x, y; i <= n - 1; ++i) {
    scanf("%d%d", &x, &y);
    add(x, y);
    add(y, x);
  }
  dfs1(1, 0, 1);
  solve(1, 1);
  for (int i = 1; i <= n; ++i) printf("%lld%c", ans[i], i == n ? '\n' : ' ');
  return 0;
}
