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;
char FREAD[100000000], *p1 = FREAD;
inline long long read() {
  char c;
  long long x = 0, w = 1;
  for (c = *p1++; c < '0' || c > '9'; c = *p1++)
    if (c == '-') w = -1;
  for (; c <= '9' && c >= '0'; c = *p1++) x = x * 10 + c - '0';
  return x * w;
}
const long long maxn = 300005;
long long n, col[maxn], h[maxn], cnt;
struct edge {
  long long to, next;
};
edge e[maxn << 1];
inline void addedge(long long x, long long y) {
  e[++cnt].next = h[x];
  e[cnt].to = y;
  h[x] = cnt;
}
long long sz[maxn], son[maxn], f[maxn], Hs[maxn];
inline void dfs(long long x) {
  sz[x] = 1;
  for (register long long i = h[x]; i; i = e[i].next) {
    long long y = e[i].to;
    if (f[x] == y) continue;
    f[y] = x;
    dfs(y);
    sz[x] += sz[y];
    if (sz[y] > sz[son[x]]) son[x] = y;
  }
}
long long Ans[maxn], sum, Mx, Son;
inline void add(long long x, long long v) {
  Hs[col[x]] += v;
  if (v == 1 && Hs[col[x]] > Mx)
    Mx = Hs[col[x]], sum = col[x];
  else if (v == 1 && Hs[col[x]] == Mx)
    sum += col[x];
  for (register long long i = h[x]; i; i = e[i].next) {
    long long y = e[i].to;
    if (f[x] == y || Son == y) continue;
    add(y, v);
  }
}
inline void dsu(long long x, long long opt) {
  for (register long long i = h[x]; i; i = e[i].next) {
    long long y = e[i].to;
    if (son[x] == y || f[x] == y) continue;
    dsu(y, 0);
  }
  if (son[x]) dsu(son[x], 1), Son = son[x];
  add(x, 1);
  Son = 0;
  Ans[x] = sum;
  if (!opt) add(x, -1), Mx = 0, sum = 0;
}
long long b[maxn], tot;
int main() {
  fread(FREAD, 1, sizeof FREAD, stdin);
  n = read();
  for (register long long i = 1; i <= n; ++i) col[i] = read();
  for (register long long i = 1; i < n; ++i) {
    long long x = read(), y = read();
    addedge(x, y);
    addedge(y, x);
  }
  dfs(1);
  dsu(1, 1);
  for (register long long i = 1; i <= n; ++i) printf("%lld ", Ans[i]);
  return 0;
}
