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;
inline int read() {
  int x = 0, f = 1;
  char s = getchar();
  for (; !isdigit(s); s = getchar())
    if (s == '-') f = -1;
  for (; isdigit(s); s = getchar()) x = x * 10 + s - 48;
  return x * f;
}
const int MAXN = 100010;
struct Edge {
  int to, nxt;
} e[MAXN << 1];
int tot, h[MAXN];
inline void add(int u, int v) {
  e[++tot] = (Edge){v, h[u]};
  h[u] = tot;
}
int n, c[MAXN], siz[MAXN], son[MAXN], vis[MAXN];
int top, cnt[MAXN];
long long sum[MAXN], ans[MAXN];
void calc(int x, int fa, int flag) {
  if (flag > 0) {
    sum[cnt[c[x]]] -= c[x];
    ++cnt[c[x]];
    sum[cnt[c[x]]] += c[x];
    if (cnt[c[x]] > top) top = cnt[c[x]];
  } else {
    sum[cnt[c[x]]] -= c[x];
    --cnt[c[x]];
    sum[cnt[c[x]]] += c[x];
    if (!sum[top] && top) --top;
  }
  for (int i = h[x]; i; i = e[i].nxt) {
    int y = e[i].to;
    if (y == fa || vis[y]) continue;
    calc(y, x, flag);
  }
}
void dfs1(int x, int fa) {
  siz[x] = 1;
  for (int i = h[x]; i; i = e[i].nxt) {
    int y = e[i].to;
    if (y == fa) continue;
    dfs1(y, x);
    siz[x] += siz[y];
    if (!son[x])
      son[x] = y;
    else if (siz[y] > siz[son[x]])
      son[x] = y;
  }
}
void dfs2(int x, int fa, int flag) {
  for (int i = h[x]; i; i = e[i].nxt) {
    int y = e[i].to;
    if (y == fa || y == son[x]) continue;
    dfs2(y, x, 0);
  }
  if (son[x]) dfs2(son[x], x, 1), vis[son[x]] = 1;
  calc(x, fa, 1);
  vis[son[x]] = 0;
  ans[x] = sum[top];
  if (!flag) calc(x, fa, -1);
}
int main() {
  n = read();
  for (int i = 1; i <= n; ++i) c[i] = read();
  for (int i = 1; i < n; ++i) {
    int u = read(), v = read();
    add(u, v), add(v, u);
  }
  dfs1(1, 0);
  dfs2(1, 0, 0);
  for (int i = 1; i <= n; ++i) printf("%lld ", ans[i]);
  return 0;
}
