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 + 5;
inline int read() {
  int x = 0;
  char c = getchar();
  while (!isdigit(c)) c = getchar();
  while (isdigit(c)) x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();
  return x;
}
struct edge {
  int to, next;
} e[N << 1];
int head[N], tot;
inline void add(int u, int v) {
  e[++tot] = (edge){v, head[u]};
  head[u] = tot;
}
int n, siz[N], son[N], col[N], Son, cnt[N], maxx, fa[N];
long long sum, ans[N];
void dfs1(int x, int ff) {
  siz[x] = 1;
  fa[x] = ff;
  for (int i = head[x]; i; i = e[i].next) {
    int v = e[i].to;
    if (v == ff) continue;
    dfs1(v, x);
    siz[x] += siz[v];
    if (siz[son[x]] < siz[v]) son[x] = v;
  }
}
void addsum(int x, int val) {
  cnt[col[x]] += val;
  if (cnt[col[x]] > maxx)
    maxx = cnt[col[x]], sum = col[x];
  else if (cnt[col[x]] == maxx)
    sum += col[x];
  for (int i = head[x]; i; i = e[i].next) {
    int v = e[i].to;
    if (v == Son || v == fa[x]) continue;
    addsum(v, val);
  }
}
void dfs2(int x, int op) {
  for (int i = head[x]; i; i = e[i].next) {
    int v = e[i].to;
    if (v == fa[x] || v == son[x]) continue;
    dfs2(v, 0);
  }
  if (son[x]) dfs2(son[x], 1), Son = son[x];
  addsum(x, 1);
  Son = 0;
  ans[x] = sum;
  if (!op) addsum(x, -1), sum = 0, maxx = 0;
}
int main() {
  n = read();
  int u, v;
  for (int i = 1; i <= n; ++i) col[i] = read();
  for (int i = 1; i < n; ++i) {
    u = read();
    v = read();
    add(u, v);
    add(v, u);
  }
  dfs1(1, 0);
  dfs2(1, 0);
  for (int i = 1; i <= n; ++i) printf("%lld ", ans[i]);
}
