Problem:
Score : 700 points

Problem Statement

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.

Sample Input 1

3 2
1 2 3

Sample Output 1

2 3

In this case, X=(1,2,3,1,2,3).
We will perform the operations as follows:

i=0: s does not contain 1, so we add 1 to the end of s, resulting in s=(1).

i=1: s does not contain 2, so we add 2 to the end of s, resulting in s=(1,2).

i=2: s does not contain 3, so we add 3 to the end of s, resulting in s=(1,2,3).

i=3: s does contain 1, so we repeatedly delete the element at the end of s as long as s contains 1, which causes the following changes to s: (1,2,3)→(1,2)→(1)→().

i=4: s does not contain 2, so we add 2 to the end of s, resulting in s=(2).

i=5: s does not contain 3, so we add 3 to the end of s, resulting in s=(2,3).

Sample Input 2

5 10
1 2 3 2 3

Sample Output 2

3

Sample Input 3

6 1000000000000
1 1 2 2 3 3

Sample Output 3

s may be empty in the end.

Sample Input 4

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

Sample Output 4

9 2 6

Solution in C (GCC 5.4.1) (judge verdict: Wrong Answer)

#include<stdio.h>
long long n,m,c=0,i,j,d[400010],f[200010],t[200010],k[200010],r=0;
long long b[200010]={};
int main(){
  scanf("%lld %lld",&n,&m);
  for(i=0;i<n;i++)scanf("%lld",&d[i]);
  for(i=0;i<n;i++)d[i+n]=d[i];
  for(i=0;i<2e5+5;i++)f[i]=t[i]=-1;
  //n*=2;
  for(i=0;i<n*2;i++){
    if(f[d[i]]==-1)f[d[i]]=i;
    else{
      t[f[d[i]]]=i;
      f[d[i]]=i;
    }
  }
  //for(i=0;i<n;i++)printf("%lld:%lld ",i,t[i]);printf("\n");
  for(i=0;1;){
    j=t[i]+1;//printf("%lld\n",d[i]);
    if(j>n){
      c++;
      k[r]=i;
      //if(j!=i+n+1)b[r]=1;
      r++;
      //printf("%lld\n",d[i]);
    }
    i=j%n;
    if(i==0)break;
  }
  //if(r==0)return 0;
  //for(i=0;i<r;i++)printf("%lld:%lld ",d[k[i]],k[i]);printf("\n");//return 0;
  j=0;
  if((m-1)%(c+1)==c)return 0;
  for(i=k[(m-1)%(c+1)];1;){
    if(t[i]>=n){
      if(j)printf(" ");
      printf("%lld",d[i]);
      i++;
      j=1;
    }
    else{
      i=t[i]+1;
    }
    if(i==n)break;
    //printf(" ");
  }
  if(j)printf("\n");
  return 0;
  /*{
    printf("%lld",k[i]);
    if(b[i]){
      printf("\n");
      break;
    }
    if(i+1<r)printf(" ");
    else printf("\n");
  }//printf("\n");
  */
  return 0;
}