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++)
using ll = long long;
const int MAX = 2e5+1;

int main(){
  ll n, k;
  cin >> n >> k;
  int a[n];
  rep(i,n) cin >> a[i];
  vector<set<int>> s(MAX);
  rep(i,n) {
    s[a[i]].insert(i);
    s[a[i]].insert(i+n);
  }
  s[a[0]].insert(2*n);
  vector<int> t(2*n+1);
  for(int i = n; i <= 2*n; i++) t[i] = i-n;
  for(int i = n-1; i >= 0; i--) {
    auto itr = s[a[i]].upper_bound(i);
    t[i] = t[*itr+1];
  }
  vector<int> p(n+1), q(n+1);
  rep(i,n+1) p[i] = i;
  ll d = k-1;
  while(d > 0) {
    if(d % 2) {
      rep(i,n+1) p[i] = t[p[i]];
    }
    rep(i,n+1) q[i] = t[t[i]];
    rep(i,n+1) t[i] = q[i];
    d /= 2;
  }
  int pos = p[0];
  vector<int> ans;
  while(pos < n) {
    auto itr = s[a[pos]].upper_bound(pos); 
    if(*itr >= n) {
      ans.push_back(a[pos]);
      pos++;
    } else {
      pos = *itr + 1;
    }
  }
  for(auto k : ans) cout << k << " ";
  cout << endl;
  return 0;
}
