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 bool EQ(double a, double b) { return fabs(a - b) < 1e-9; }
inline void normal(long long int &a) {
  a %= 1000000007;
  (a < 0) && (a += 1000000007);
}
inline long long int modMul(long long int a, long long int b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  return (a * b) % 1000000007;
}
inline long long int modAdd(long long int a, long long int b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  return (a + b) % 1000000007;
}
inline long long int modSub(long long int a, long long int b) {
  a %= 1000000007, b %= 1000000007;
  normal(a), normal(b);
  a -= b;
  normal(a);
  return a;
}
inline long long int modPow(long long int b, long long int p) {
  long long int r = 1;
  while (p) {
    if (p & 1) r = modMul(r, b);
    b = modMul(b, b);
    p >>= 1;
  }
  return r;
}
inline long long int modInverse(long long int a) {
  return modPow(a, 1000000007 - 2);
}
inline long long int modDiv(long long int a, long long int b) {
  return modMul(a, modInverse(b));
}
inline bool isInside(pair<int, int> p, long long int n, long long int m) {
  return (p.first >= 0 && p.first < n && p.second >= 0 && p.second < m);
}
inline bool isInside(pair<int, int> p, long long int n) {
  return (p.first >= 0 && p.first < n && p.second >= 0 && p.second < n);
}
inline bool isSquare(long long int x) {
  long long int s = sqrt(x);
  return (s * s == x);
}
inline bool isFib(long long int x) {
  return isSquare(5 * x * x + 4) || isSquare(5 * x * x - 4);
}
inline bool isPowerOfTwo(long long int x) {
  return ((1LL << (long long int)log2(x)) == x);
}
template <typename first, typename second>
ostream &operator<<(ostream &os, const pair<first, second> &p) {
  return os << "(" << p.first << ", " << p.second << ")";
}
template <typename T>
ostream &operator<<(ostream &os, const vector<T> &v) {
  os << "{";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "}";
}
template <typename T>
ostream &operator<<(ostream &os, const set<T> &v) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << *it;
  }
  return os << "]";
}
template <typename first, typename second>
ostream &operator<<(ostream &os, const map<first, second> &v) {
  os << "[";
  for (auto it = v.begin(); it != v.end(); ++it) {
    if (it != v.begin()) os << ", ";
    os << it->first << " = " << it->second;
  }
  return os << "]";
}
clock_t tStart = clock();
void faltu() { cerr << endl; }
template <typename T>
void faltu(T a[], int n) {
  for (int i = 0; i < n; ++i) cerr << a[i] << ' ';
  cerr << endl;
}
template <typename T, typename... hello>
void faltu(T arg, const hello &...rest) {
  cerr << arg << ' ';
  faltu(rest...);
}
struct func {
  bool operator()(pair<int, int> const &a, pair<int, int> const &b) {
    if (a.first == b.first) return (a.second < b.second);
    return (a.first < b.first);
  }
};
const long long int INF = 0x3f3f3f3f3f3f3f3f;
const long double EPS = 1e-9;
const int inf = 0x3f3f3f3f;
const int N = (int)2e5 + 9;
int n, i, c[N], w[N], used[N], mx[N];
map<int, int> second[N];
long long int cnt[N];
vector<int> g[N];
long long int ans[N];
void run_merge(int a, int b) {
  if (second[w[a]].size() < second[w[b]].size()) {
    swap(w[a], w[b]);
  }
  int id = w[a];
  map<int, int>::iterator it;
  for (it = second[w[b]].begin(); it != second[w[b]].end(); it++) {
    int val = (*it).first;
    int amount = (*it).second;
    second[id][val] += amount;
    if (second[id][val] > mx[id]) {
      mx[id] = second[id][val];
      cnt[id] = 0;
    }
    if (second[id][val] == mx[id]) {
      cnt[id] += val;
    }
  }
}
void dfs(int v) {
  used[v] = 1;
  for (int i = 0; i < g[v].size(); i++) {
    int to = g[v][i];
    if (used[to]) continue;
    dfs(to);
    run_merge(v, to);
  }
  ans[v] = cnt[w[v]];
}
int main() {
  ios_base::sync_with_stdio(false);
  ;
  cin >> n;
  for (int i = 1; i <= (n); i++) {
    cin >> c[i];
    second[i][c[i]] = 1;
    w[i] = i;
    mx[i] = 1;
    cnt[i] = c[i];
  }
  for (int i = 1; i <= (n - 1); i++) {
    int a, b;
    cin >> a >> b;
    g[a].push_back(b);
    g[b].push_back(a);
  }
  dfs(1);
  for (int i = 1; i <= (n); i++) cout << ans[i] << ' ';
}
