We have a sequence of N \times K integers: X=(X_0,X_1,\cdots,X_{N \times K-1}). Its elements are represented by another sequence of N integers: A=(A_0,A_1,\cdots,A_{N-1}). For each pair i, j (0 \leq i \leq K-1,\ 0 \leq j \leq N-1), X_{i \times N + j}=A_j holds.

Snuke has an integer sequence s, which is initially empty. For each i=0,1,2,\cdots,N \times K-1, in this order, he will perform the following operation:

* If s does not contain X_i: add X_i to the end of s.
* If s does contain X_i: repeatedly delete the element at the end of s until s no longer contains X_i. Note that, in this case, we do not add X_i to the end of s.



Find the elements of s after Snuke finished the operations.

Constraints

* 1 \leq N \leq 2 \times 10^5
* 1 \leq K \leq 10^{12}
* 1 \leq A_i \leq 2 \times 10^5
* All values in input are integers.

Input

Input is given from Standard Input in the following format:


N K
A_0 A_1 \cdots A_{N-1}


Output

Print the elements of s after Snuke finished the operations, in order from beginning to end, with spaces in between.

Examples

Input

3 2
1 2 3


Output

2 3


Input

5 10
1 2 3 2 3


Output

3


Input

6 1000000000000
1 1 2 2 3 3


Output




Input

11 97
3 1 4 1 5 9 2 6 5 3 5


Output

9 2 6
#include <bits/stdc++.h>
using namespace std;
#define rep(i, n) for(int i = 0; i < (int)(n); i++)
#define all(x) (x).begin(), (x).end()
typedef long long ll;
static const int INF (1<<30);

int main() {
  ll N;
  ll K;
  cin >> N >> K;
  vector<pair<int,int>> A(N);
  vector<int> cnt(200005, 0);
  vector<vector<int>> id(200005);

  rep(i, N) {
    int x; cin >> x;
    A[i].first = x;
    cnt[x]++;
    A[i].second = cnt[x];
    id[x].push_back(i);
  }

  ll T = 0LL;
  int k = 0;
  vector<int> sub;
  while (true) {
    int x = A[k].first;

    if (A[k].second == cnt[x]) {
      sub.push_back(k);
      T += (ll)(N + id[x][0] - k + 1);
      k = id[x][0] + 1;
    } else {
      T += (ll)(id[x][A[k].second] - id[x][A[k].second-1] + 1);
      k = id[x][A[k].second] + 1;
    }

    if (k == N) break;
  }
  sub.push_back(N-1);

  
  ll t = T / N;
  int mod = K % t;

  if (mod == 0) {
    cout << "" << endl;
  } else {

    int k = sub[mod-1];
    while (true) {
      int x = A[k].first;
      if (A[k].second == cnt[x]) {
        cout << x << " ";
        k++;
      } else {
        k = id[x][A[k].second]+1;
      }

      if (k == N) break;
    }

    cout << endl;
  }
    

  return 0;
}