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;
int sum[10000005], rt[10000005], n, a[200005], nxt[200005], lnk[200005],
    son[200005], tot, lc[10000005], rc[10000005];
long long Mx[10000005], x, y, Ans[200005];
inline int read() {
  int ret = 0, f = 1;
  char ch = getchar();
  while (ch < '0' || ch > '9') {
    if (ch == '-') f = -f;
    ch = getchar();
  }
  while (ch <= '9' && ch >= '0') ret = ret * 10 + ch - '0', ch = getchar();
  return ret * f;
}
inline void add(int x, int y) {
  nxt[++tot] = lnk[x];
  lnk[x] = tot;
  son[tot] = y;
}
inline void push_down(int k) {
  if (sum[lc[k]] > sum[rc[k]])
    sum[k] = sum[lc[k]], Mx[k] = Mx[lc[k]];
  else if (sum[lc[k]] < sum[rc[k]])
    sum[k] = sum[rc[k]], Mx[k] = Mx[rc[k]];
  else
    sum[k] = sum[rc[k]], Mx[k] = Mx[rc[k]] + Mx[lc[k]];
}
inline void change(int &k, int L, int R) {
  if (!k) k = ++tot;
  if (L == R) {
    sum[k]++, Mx[k] = x;
    return;
  }
  int mid = L + R >> 1;
  if (x <= mid)
    change(lc[k], L, mid);
  else
    change(rc[k], mid + 1, R);
  push_down(k);
}
inline int Merge(int x, int y, int L, int R) {
  if (!x || !y) return x ^ y;
  if (L == R) {
    sum[x] += sum[y];
    return x;
  }
  int mid = L + R >> 1;
  lc[x] = Merge(lc[x], lc[y], L, mid);
  rc[x] = Merge(rc[x], rc[y], mid + 1, R);
  push_down(x);
  return x;
}
inline void Get(int x, int fa) {
  for (int i = lnk[x]; i; i = nxt[i])
    if (son[i] != fa) Get(son[i], x), rt[x] = Merge(rt[x], rt[son[i]], 1, n);
  if (sum[rt[x]] > 0) Ans[x] = Mx[rt[x]];
}
int main() {
  n = read();
  for (int i = 1; i <= n; i++) x = read(), change(rt[i], 1, n);
  tot = 0;
  for (int i = 1; i < n; i++) x = read(), y = read(), add(x, y), add(y, x);
  Get(1, 0);
  for (int i = 1; i <= n; i++) printf("%lld ", Ans[i]);
  return 0;
}
