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 <cstdio>
#include <vector>
#include <algorithm>
using namespace std;

int main()
{
  int n;
  long long K;
  scanf("%d%lld",&n,&K);
  vector<int> a(n);
  int mx=0;
  for(int i=0;i<n;i++){
    scanf("%d",&a[i]);
    mx=max(mx,a[i]);
  }
  vector<int> b(mx+1,-1),nx(2*n);
  for(int i=2*n-1;i>=0;i--){
    if(b[a[i%n]]!=-1){
      nx[i]=b[a[i%n]];
    }
    b[a[i%n]]=i;
  }
  vector<int> V(n,-1);
  V[0]=0;
  int cur=0,t=0;
  K--;
  while(1){
    int next=nx[cur];
    //printf("%d %d %d\n",cur,next,t);
    if(nx[cur]>=n){
      if(K==0){
	break;
      }
      K--;
      t++;
      next-=n;
    }
    cur=next+1;
    if(cur==n){
      if(K==0){
	break;
      }
      cur=0;
      K--;
      t++;
    }
    if(V[cur]==-1){
      V[cur]=t;
    }
    else{
      K=K%(t-V[cur]);
    }
  }
  vector<int> ans;
  while(cur<n){
    if(nx[cur]>=n){
      ans.push_back(a[cur]);
      cur++;
    }
    else{
      cur=nx[cur]+1;
    }
  }
  for(int i=0;i<ans.size();i++){
    printf("%d%c",ans[i],i+1==ans.size()?'\n':' ');
  }
  return 0;
}
