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;
template <typename T>
inline void read(T &x) {
  char ch = getchar();
  bool flag = false;
  x = 0;
  while (!isdigit(ch)) flag |= (ch == '-'), ch = getchar();
  while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
  if (flag) x = -x;
}
const int inf = 0x3f3f3f3f;
const double eps = 10e-5;
const int Mod = 1e9 + 7;
const int N = 1e5 + 5, M = 105;
struct Edge {
  int to, next;
} e[N * 2];
int head[N], c[N], cnt[N], sz[N];
int n, tot, son[N], Bigson;
long long ans[N], maxn, Sum;
void add(int x, int y) {
  e[++tot].to = y;
  e[tot].next = head[x];
  head[x] = tot;
}
void dfs1(int u, int fa) {
  sz[u] = 1;
  for (int i = head[u]; ~i; i = e[i].next) {
    int &v = e[i].to;
    if (v == fa) continue;
    dfs1(v, u);
    sz[u] += sz[v];
    if (sz[v] > sz[son[u]]) son[u] = v;
  }
}
void update(int u, int fa, int val) {
  cnt[c[u]] += val;
  if (cnt[c[u]] > maxn)
    maxn = cnt[c[u]], Sum = c[u];
  else if (cnt[c[u]] == maxn)
    Sum += c[u];
  for (int i = head[u]; ~i; i = e[i].next) {
    int &v = e[i].to;
    if (v == fa || v == Bigson) continue;
    update(v, u, val);
  }
}
void dfs2(int u, int fa, int op) {
  for (int i = head[u]; ~i; i = e[i].next) {
    int &v = e[i].to;
    if (v == fa || v == son[u]) continue;
    dfs2(v, u, 0);
  }
  if (son[u]) dfs2(son[u], u, 1), Bigson = son[u];
  update(u, fa, 1);
  ans[u] = Sum, Bigson = 0;
  if (!op) {
    update(u, fa, -1);
    Sum = maxn = 0;
  }
}
int main() {
  memset(head, -1, sizeof(head));
  read(n);
  for (int i = 1; i <= n; i++) read(c[i]);
  for (int i = 1; i < n; i++) {
    int x, y;
    read(x), read(y);
    add(x, y);
    add(y, x);
  }
  dfs1(1, 0);
  dfs2(1, 0, 0);
  for (int i = 1; i <= n; i++) printf("%lld%c", ans[i], i != n ? ' ' : '\n');
  return 0;
}
