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;

const long MOD=1000000007;
const long INF=100000000000000000;
int main(){
    long N,K;
    cin>>N>>K;
    vector<long> A(N);
    for(int i=0;i<N;i++){
        cin>>A.at(i);
    }
    
    vector<long> next(N);//次にその値がくる点なければ-1;
    vector<long> front(200010,-1);//計算のため。A.at(i)のそこまでの先頭
    for(int i=N-1;i>=0;i--){
        next.at(i)=front.at(A.at(i));
        front.at(A.at(i))=i;
    }
    
    vector<long> pass(N+1);//数列Aをiばんめから始めた時に先頭になる点の添字。次ならNとなる。
    pass.at(N)=N;
    for(int i=N-1;i>=0;i--){
        if(next.at(i)==-1) pass.at(i)=i;
        else pass.at(i)=pass.at(next.at(i)+1);
    }
    
    long syuki=1,prei=pass.at(0);
    while(prei!=N){
        syuki++;
        prei=pass.at(front.at(A.at(prei))+1);
    }
    
    //cout<<syuki<<endl;
    
    
    K%=syuki;
    long pos=0;
    //cout<<K<<endl;
    while(K-1>0){//最後は出力と同時にシミュレートする
        K--;
        pos=front.at(A.at(pass.at(pos)))+1;
    }
    //cout<<pos<<endl;
    long para=pos;
    while(para<N){
        if(next.at(para)==-1){
            cout<<A.at(para)<<" ";
            para++;
        }else{
            para=next.at(para)+1;
        }
    }
    cout<<endl;
}