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>
#include<stdlib.h>
#include<string.h>
#define df 0
#define MAX 200010

typedef long int li;
int n;
li k;
li orbit[MAX]={};
li accum[MAX]={};
li len_orbit=0;
li term=0;
li* map[MAX];
li deg[MAX]={};
int a[MAX];


int next_orbit(int i);
int search(int t,int i);
int last_orbit(int t);

int main(){
  scanf("%d%d",&n,&k);
  int i;
  for(i=1;i<=n;i++) map[i]=NULL;
  for(i=0;i<n;i++){
    scanf("%d",&a[i]);
    deg[a[i]]++;
    map[a[i]]=realloc(map[a[i]],sizeof(li)*deg[a[i]]);
    map[a[i]][deg[a[i]]-1]=i;
  }



  orbit[0]=0;len_orbit=1;
  i=0;
  do{
    if(df)printf("i:%d\n",i);
    i=next_orbit(i);
  }while(i!=0);
  
  if(df){
    printf("len:%d\n",len_orbit);
    for(i=0;i<len_orbit;i++){
      printf(" %d",orbit[i]);
    }
    printf("\n");
  }
  k%=term;
  if(k==0)return 0;
  i=last_orbit(k);
  if(df)printf("last:%d\n",i);
  while(i<n){
    int j=search(a[i],i);
    if(j==deg[a[i]]-1){
      printf("%d ", a[i]);
      i++;
    }else{
      i=map[a[i]][j+1]+1;
    }
  }

  return 0;
}

int next_orbit(int i){
  int j=search(a[i],i);
  if(df)printf("j:%d->",j,deg[a[i]]);
  if(j==deg[a[i]]-1){
    term++;
    j=0;
  }else j++;
  if(df)printf("j:%d,",j);
  i=map[a[i]][j]+1;
  if(i==n){term++; i=0;}
  if(df)printf("i:%d\n",i);
  accum[len_orbit]=term+1;
  orbit[len_orbit++]=i;
  return i;
}

int search(int t,int i){
  int a=0,b=deg[t]-1;
  if(map[t][b]==i)return b;
  while(b-a>0){
    int c=(a+b)/2;
    if(map[t][c]==i)return c;
    if(map[t][c]<i) a=c;
    else b=c;
  }
  return a;
}


int last_orbit(int t){
  int a=0,b=len_orbit-1;
  while(b-a>1){
    int c=(a+b)/2;
    if(accum[c]<= t)a=c;
    else b=c;
  }
  return orbit[a];
}



/// confirm df==0 ///