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: Runtime Error)

#include <stdio.h>
#define MAX 200001

int main(){
  int n,k,a[MAX],s[MAX],list[MAX],round[MAX],next[MAX],last[MAX],count,p,mod,
      i,j;
  
  scanf("%d%d",&n,&k);
  for(i=0;i<MAX;i++){last[i]=-1;}
  for(i=0;i<n;i++){
    scanf("%d",&a[i]);
    if(last[a[i]] == -1){
      next[i] = i;
      last[a[i]] = i;
    }else{
      next[i] = next[last[a[i]]];
      next[last[a[i]]] = i;
      last[a[i]] = i;
    }
  }
  
  p=0;
  count=1;
  round[1] = 0;
  while(1){
    if(next[p] <= p){
      count++;
    }
    p = next[p] + 1;
    if(p == n){
      break;
    }
    round[count] = p;
  }

  mod = k % count;
  if(mod == 0){
    return 0;
  }
  
  p = 0;
  for(i=0;i<MAX;i++){s[i]=-1; list[i]=-1;}
  for(i=round[mod];i<n;i++){
    if(list[a[i]] == -1){
      s[p] = a[i];
      list[a[i]] = p;
      p++;
    }else{
      for(j=list[a[i]]+1;j<p;j++){
        list[s[j]] = -1;
      }
      p = list[a[i]];
      list[a[i]] = -1;
    }
  }
    
  for(i=0;i<p;i++) printf("%d ",s[i]);
  return 0;
}