Problem: You are given an array s consisting of n integers.

You have to find any array t of length k such that you can cut out maximum number of copies of array t from array s.

Cutting out the copy of t means that for each element t_i of array t you have to find t_i in s and remove it from s. If for some t_i you cannot find such element in s, then you cannot cut out one more copy of t. The both arrays can contain duplicate elements.

For example, if s = [1, 2, 3, 2, 4, 3, 1] and k = 3 then one of the possible answers is t = [1, 2, 3]. This array t can be cut out 2 times. 

  * To cut out the first copy of t you can use the elements [1, \underline{2}, 3, 2, 4, \underline{3}, \underline{1}] (use the highlighted elements). After cutting out the first copy of t the array s can look like [1, 3, 2, 4]. 
  * To cut out the second copy of t you can use the elements [\underline{1}, \underline{3}, \underline{2}, 4]. After cutting out the second copy of t the array s will be [4]. 



Your task is to find such array t that you can cut out the copy of t from s maximum number of times. If there are multiple answers, you may choose any of them.

Input

The first line of the input contains two integers n and k (1 ≤ k ≤ n ≤ 2 ⋅ 10^5) — the number of elements in s and the desired number of elements in t, respectively.

The second line of the input contains exactly n integers s_1, s_2, ..., s_n (1 ≤ s_i ≤ 2 ⋅ 10^5).

Output

Print k integers — the elements of array t such that you can cut out maximum possible number of copies of this array from s. If there are multiple answers, print any of them. The required array t can contain duplicate elements. All the elements of t (t_1, t_2, ..., t_k) should satisfy the following condition: 1 ≤ t_i ≤ 2 ⋅ 10^5.

Examples

Input


7 3
1 2 3 2 4 3 1


Output


1 2 3 


Input


10 4
1 3 1 3 10 3 7 7 12 3


Output


7 3 1 3


Input


15 2
1 2 1 1 1 2 1 1 2 1 2 1 1 1 1


Output


1 1 

Note

The first example is described in the problem statement.

In the second example the only answer is [7, 3, 1, 3] and any its permutations. It can be shown that you cannot choose any other array such that the maximum number of copies you can cut out would be equal to 2.

In the third example the array t can be cut out 5 times.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
namespace forever {
template <class type1, class type2>
inline istream &operator>>(istream &in, pair<type1, type2> &p) {
  in >> p.first >> p.second;
  return in;
}
template <class type1, class type2>
inline ostream &operator<<(ostream &o, pair<type1, type2> &p) {
  o << "(" << p.first << "," << p.second << ") ";
  return o;
}
template <class type>
inline ostream &operator<<(ostream &o, vector<type> &v) {
  for (long long i = 0; i < v.size(); i++) o << v[i] << " ";
  return o;
}
template <class type>
inline ostream &operator<<(ostream &o, set<type> &v) {
  for (long long x : v) o << x << " ";
  return o;
}
template <class t>
inline istream &operator>>(istream &in, vector<t> &p) {
  for (auto &x : p) in >> x;
  return in;
}
long long power(long long x, long long y, long long p = 1000000007) {
  long long res = 1;
  x %= p;
  while (y > 0) {
    if (y & 1) res = (res * x) % p;
    y >>= 1;
    x = (x * x) % p;
  }
  return res;
}
struct custom_hash {
  static uint64_t splitmix64(uint64_t x) {
    x += 0x9e3779b97f4a7c15;
    x = (x ^ (x >> 30)) * 0xbf58476d1ce4e5b9;
    x = (x ^ (x >> 27)) * 0x94d049bb133111eb;
    return x ^ (x >> 31);
  }
  size_t operator()(pair<uint64_t, uint64_t> x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x.first + FIXED_RANDOM) ^
           (splitmix64(x.second + FIXED_RANDOM) >> 1);
  }
  size_t operator()(uint64_t x) const {
    static const uint64_t FIXED_RANDOM =
        chrono::steady_clock::now().time_since_epoch().count();
    return splitmix64(x + FIXED_RANDOM);
  }
};
}  // namespace forever
using namespace forever;
void test_cases() {
  long long n, i, k, sum = 0;
  cin >> n >> k;
  vector<long long> a(n);
  cin >> a;
  long long s[1 << 18] = {0};
  for (long long x : a) {
    s[x]++;
  }
  vector<long long> ans;
  long long mid, l = 1, r = n;
  while (l <= r) {
    mid = (l + r) >> 1;
    vector<long long> te;
    for (i = 0; i < 200001; i++) {
      long long temp = (s[i] / mid);
      long long tempo = (k - te.size());
      temp = min(temp, tempo);
      for (long long j = 0; j < temp; j++) {
        te.push_back(i);
      }
    }
    if (te.size() == k) {
      l = mid + 1;
      ans.clear();
      ans = te;
    } else {
      r = mid - 1;
    }
  }
  cout << ans << "\n";
}
int32_t main() {
  ios_base::sync_with_stdio(false);
  cin.tie(NULL);
  cout.tie(NULL);
  long long t = 1;
  while (t--) {
    test_cases();
  }
  return 0;
}