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++14 (GCC 5.4.1) (judge verdict: Accepted)

#include<iostream>
#include<vector>
#include<map>

class C
{
  int N;
  long long K;
  std::vector<int> A;
public:
  void input();
  void solve();
};

void C::input(){
  std::cin>>N>>K;
  A.resize(N);
  for(int i=0;i<N;i++){
    std::cin>>A[i];
  }
}

void C::solve(){
  C::input();
  
  const int LOG=44;
  for(int i=0;i<N;i++) A.push_back(A[i]);
  std::map<int,int> mp;
  // next[i][j]:jからスタートし、Kを2^i回行った後に空でなくなるときのindex
  int next[LOG][N+1];
  for(int i=2*N-1;i>=N;i--){
    mp[A[i]]=i;
  }
  for(int i=N-1;i>=0;i--){
    int x=mp[A[i]]+1;
    if(x>=N){
      next[0][i]=x-N;
    }else{
      next[0][i]=next[0][x];
    }
    mp[A[i]]=i;
  }
  
  next[0][N]=0;
  for(int i=1;i<LOG;i++){
    for(int j=0;j<=N;j++){
      next[i][j]=next[i-1][next[i-1][j]];
    }
  }

  int cur=0;
  K--;
  for(int i=0;i<LOG;i++){
    if((K>>i)&1){
      cur=next[i][cur];
    }
  }

  std::vector<int> ans;
  mp.clear();
  for(int i=cur;i<N;i++){
    if(mp.count(A[i])){
      while(ans.back()!=A[i]){
        mp.erase(ans.back());
        ans.pop_back();
      }
      mp.erase(A[i]);
      ans.pop_back();
    }else{
      ans.push_back(A[i]);
      mp[A[i]]=1;
    }
  }
  for(int a:ans) std::cout<<a<<" ";
  std::cout<<"\n";
  return;
}

int main(){
  C c;
  c.solve();
  return 0;
}